AP Calculus BC: 5.3 Determining Intervals on Which a Function Is  Increasing or Decreasing – Exam Style questions with Answer- FRQ

Question

Fish enter a lake at a rate modeled by the function E given by \(E(t)=20+15 sin(\frac{\pi t}{6}).\) Fish leave the lake at a rate modeled by the function L given by \(L(t)=4+2^{0.1t^{2}}.\) Both E(t) and L (t) are measured in fish per hour, and t is measured in hours since midnight (t = 0).
(a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ? Give your answer to the nearest whole number.
(b) What is the average number of fish that leave the lake per hour over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ?
(c) At what time t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake? Justify your answer.
(d) Is the rate of change in the number of fish in the lake increasing or decreasing at 5 A.M. (t = 5) ? Explain your reasoning. 

Answer/Explanation

Ans:

(a) \(\int_{0}^{5}E(t)dt=153.457690\)

To the nearest whole number, 153 fish enter the lake from midnight to 5 A.M.

(b) \(\frac{1}{5-0}\int_{0}^{5}L(t)dt=6.059038\)

The average number of fish that leave the lake per hour from midnight to 5 A.M. is 6.059 fish per hour.

(c) The rate of change in the number of fish in the lake at time t is given by E(t)-L(t) .
E(t)-L(t) =0⇒ t = 6.20356
E(t) – L(t) > 0 for 0 ≤ t < 6.20356, and E(t) – L(t) < 0 for 6.20356 < t ≤ 8. Therefore the greatest number of fish in the lake is at time 6.204 (or 6.203).
— OR —
Let A (t) be the change in the number of fish in the lake from midnight to t hours after midnight.

\(A(t)=\int_{0}^{t}(E(s)-L(s))ds\)

A'(t) = E(t) – L(t) = 0 ⇒  t = C = 6.20356

tA(t)

0

C

8

0

135.01492

80.91998

Therefore the greatest number of fish in the lake is at time t = 6.204 (or 6.203).

(d)  E'(5) – L'(5) = -10.7228 < 0

Because E ‘ (5) – L’ (5) < 0, the rate of change in the number of fish is decreasing at time  t = 5.

Question

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for −5 ≤ x < 3, and g (x) = 2(x −4 ) 2 for 3 ≤ x ≤ 6.
(a) If f (1) = 3, what is the value of f(−5) ?
(b) Evaluate \(\int_{1}^{6}g(x)dx.\) 
(c) For −5 < x < 6, on what open intervals, if any, is the graph of f both increasing and concave up? Give a reason for your answer.
(d) Find the x-coordinate of each point of inflection of the graph of f. Give a reason for your answer.

Answer/Explanation

Ans:

(a) \(f(-5)=f(1)+\int_{1}^{-5}g(x)dx=f(1)-\int_{-5}^{1}g(x)dx\)

\(=3-\left ( -9-\frac{3}{2}+1 \right )=3-\left ( -\frac{19}{2} \right )=\frac{25}{2}\)

(b) \(\int_{1}^{6}g(x)dx=\int_{1}^{3}g(x)dx+\int_{3}^{6}g(x)dx\)

\(=\int_{1}^{3}2dx+\int_{3}^{6}2(x-4)^{2}dx\)

\(=4+_{x=3}\left [ \frac{2}{3}(x-4)^{3} \right ]^{x=6}=4+\frac{16}{3}-\left ( -\frac{2}{3} \right )=10\)

(c) The graph of f is increasing and concave up on 0 < x < 1 and 4 < x < 6 because f ‘(x) =g(x)> 0 and f ‘(x) = g(x)  is increasing on those intervals.

(d) The graph of f has a point of inflection at x = 4 because f'(x) = g(x) changes from decreasing to increasing at x = 4.

Question

Let f be the function defined by \(f(x)=\frac{3}{2x^{2}-7x+5}.\) 
(a) Find the slope of the line tangent to the graph of f at x = 3.
(b) Find the x-coordinate of each critical point of f in the interval 1 < x < 2.5. Classify each critical point as the location of a relative minimum, a relative maximum, or neither. Justify your answers.
(c) Using the identity that \(\frac{3}{2x^{2}-7x+5}=\frac{2}{2x-5}-\frac{1}{x-1}, evaluate\int_{5}^{\infty }f(x)\) or show that the integral diverges. 
(d) Determine whether the series \(\sum_{n=5}^{\infty }\frac{3}{2n^{2}-7n+5}\) converges or diverges. State the conditions of the test used for determining convergence or divergence.

Answer/Explanation

Ans:

(a) \(f'(x)=\frac{-3(4x-7)}{\left ( 2x^{2}-7x+5 \right )^{2}}\)

\(f'(3)=\frac{-3(5)}{\left ( 18-21+5 \right )^{2}}=-\frac{15}{4}\)

(b) \(f'(x)=\frac{-3(4x-7)}{\left ( 2x^{2}-7x+5 \right )^{2}}=0\Rightarrow x=\frac{7}{4}\)

The only critical point in the interval 1 < x < 2.5  has x-coordinate \(\frac{7}{4}.\) f ′ changes sign from positive to negative at \(x=\frac{7}{4}.\) Therefore, f has a relative maximum at \(x=\frac{7}{4}.\) 

(c) \(\int_{5}^{\infty }f(x)dx=\lim_{b\rightarrow \infty }\int_{5}^{b}\frac{3}{2x^{2}-7x+5}dx =\lim_{b\rightarrow \infty }\int_{5}^{b}\left ( \frac{2}{2x-5} -\frac{1}{x-1}\right )dx\)

\(=\lim_{b\rightarrow \infty }\left [ In(2x-5)-In(x-1) \right ]_{5}^{b}=\lim_{b\rightarrow \infty }\left [ In\left ( \frac{2x-5}{x-1} \right ) \right ]_{5}^{b}\)

\(=\lim_{b\rightarrow \infty }\left [ In(\frac{2b-5}{b-1})-In(\frac{5}{4}) \right ]=In 2-In\left ( \frac{5}{4} \right )=In\left ( \frac{8}{5} \right )\)

(d) f is continuous, positive, and decreasing on [5, ∞). The series converges by the integral test since \(\int_{5}^{\infty }\frac{3}{2x^{2}-7x+5}dx\) converges.

-OR –

\(\frac{3}{2n^{2}-7n+5}>0 and \frac{1}{n^{2}}>0 for n\geq 5.\)

Since \(\frac{\frac{3}{2n^{2}-7n+5}}{\frac{1}{n^{2}}} =\frac{3}{2}\) and the series \(\sum_{n=5}^{\infty }\frac{1}{n^{2}}\) converges, the series \(\sum_{n=5}^{\infty }\frac{3}{2n^{2}-7n+5}\) converges by the limit comparison test.

Question

The function f is differentiable on the closed interval [−6, 5 ] and satisfies f (−2 )2 = 7. The graph of f ‘, the derivative of f, consists of a semicircle and three line segments, as shown in the figure above.
(a) Find the values of f (−6) and f (5).
(b) On what intervals is f increasing? Justify your answer.
(c) Find the absolute minimum value of f on the closed interval [−6, 5 ]. Justify your answer.
(d) For each of f “(−5) and f “(3), find the value or explain why it does not exist.

Answer/Explanation

Ans:

(a) \(f(-6)=f(-2)+\int_{-2}^{-6}f'(x)dx=7-\int_{-6}^{-2}f'(x)dx=7-4=3\)

\(f(5)=f(-2)+\int_{-2}^{5}f'(x)dx=7-2\pi +3=10-2\pi \)

(b) f ′ ( x) > 0 on the intervals [− 6, -2) and (2, 5). Therefore, f is increasing on the intervals [− 6, -2] and [2, 5] .

(c) The absolute minimum will occur at a critical point where f ‘ (x) = 0 or at an endpoint.
f ′( x) = 0 ⇒x = – 2, x = 2

xf(x)

-6

-2

2

5

3

7

7-2π

10-2π

The absolute minimum value is f (2) = 7 − 2π

(d) \(f”(-5)=\frac{2-0}{-6-(-2)}=-\frac{1}{2}\)

\(\lim_{x\rightarrow 3^{-}}\frac{f'(x)-f'(3)}{x-3}=2 and \lim_{x\rightarrow 3^{+}}\frac{f'(x)-f'(3)}{x-3}=-1\)

f ′′(3) does not exist because

\(\lim_{x\rightarrow 3^{-}}\frac{f'(x)-f'(3)}{x-3}\neq \lim_{x\rightarrow 3^{+}}\frac{f'(x)-f'(3)}{x-3}.\)

Question

The figure above shows the polar curves r = f (θ) q = 1+ sin θ cos  (2θ) and r = g(θ)  = 2 cos θ for \(\(0 \leq \theta \leq \frac{\pi }{2}.\)\)  Let R be the region in the first quadrant bounded by the curve r = f (θ)  and the x-axis. Let S be the region in the first quadrant bounded by the curve r = f (θ) , the curve r = g (θ), and the x-axis.
(a) Find the area of R.
(b) The ray θ = k, where \(0 < k < \frac{\pi }{2},\) divides S into two regions of equal area. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of k.
(c) For each θ, \(0 \leq \theta \leq \frac{\pi }{2},\) let w(θ) be the distance between the points with polar coordinates (f(θ), θ) and (g(θ), θ). Write an expression for w(θ). Find wA,  the average value of w(θ) over the interval \(0 \leq \theta \leq \frac{\pi }{2}.\) 
(d) Using the information from part (c), find the value of θ for which w(θ) = wA. Is the function w (θ) increasing or decreasing at that value of θ ? Give a reason for your answer.

Answer/Explanation

Ans:

(a) \(\frac{1}{2}\int_{0}^{\pi /2}(f(\theta ))^{2}d\theta =0.648414\)
The area of R is 0.648.

(b) \(\int_{0}^{k}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta =\frac{1}{2}\int_{0}^{\pi /2}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta\)

-OR-

\(\int_{0}^{k}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta =\int_{0}^{\pi /2}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta\)

(c) w(θ) = g(θ) – f(θ)

\(w_{A}=\frac{\int_{0}^{\pi /2}w(\theta )d\theta }{\frac{\pi }{2}-0}=0.485446\)

The average value of w(θ ) on the interval \(\left [ 0, \frac{\pi }{2} \right ]\) is 0.485.

(d) \(w(\theta )=w_{A}for 0\leq \theta \leq \frac{\pi }{2}\Rightarrow \theta =0.517688\)

w(θ) = wA at θ = 0.518 (or 0.517).

w'(0.518) < 0 ⇒ w(θ) is decreasing at θ = 0.518.

Question

The figure above shows the graph of the piecewise-linear function f. For -4 ≤ x ≤ 12, the function g is defined by \(g(x)=\int_{2}^{x}f(t)dt.\) 
(a) Does g have a relative minimum, a relative maximum, or neither at x = 10 ? Justify your answer.
(b) Does the graph of g have a point of inflection at x = 4 ? Justify your answer.
(c) Find the absolute minimum value and the absolute maximum value of g on the interval -4 ≤ x ≤ 12. Justify your answers.
(d) For -4 ≤ x ≤ 12, find all intervals for which g(x) ≤ 0.

Answer/Explanation

Ans:

(a) The function g has neither a relative minimum nor a relative maximum at x = 10 since g′ (x)= f(x) and f (x) ≤ 0 for 8 ≤ x ≤ 12.

(b) The graph of g has a point of inflection at x = 4 since g′(x) = f(x) is increasing for 2 ≤ x ≤  4 and decreasing for 4 ≤ x ≤  8.

(c) g′(x) = f(x) changes sign only at x = −2 and x = 6.

xg(x)

-4

-2

6

12

-4

-8

8

-4

On the interval −4 ≤ x ≤ 12, the absolute minimum value is g(−2) =− 8  and the absolute maximum value is g(6) = 8.

(d) g(x) ≤ 0 for −4 ≤ x ≤ 2 and 10 ≤ x ≤  12.

Question

The rate at which rainwater flows into a drainpipe is modeled by the function R, where \(R(t)=20 sin \left ( \frac{t^{2}}{35} \right )\) cubic feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. out the other end of the pipe at a rate modeled by D(t) = -0.04t3 + 0.4t2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8. There are 30 cubic feet of water in the pipe at time t = 0.
(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8 ?
(b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your answer.
(c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer.
(d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the pipe will begin to overflow. 

Answer/Explanation

Ans:

(a) \(\int_{0}^{8}R(t)dt=76.570\)

(b) R(3) – D(3) = -0.313632 < 0
Since R(3) < D(3), the amount of water in the pipe is decreasing at time t  = 3 hours.

(c) The amount of water in the pipe at time t, 0 ≤ t ≤ 8, is 

\(30+\int_{0}^{t}\left [ R(x)-D(x) \right ]dx.\)

R(t) – D(t) = 0 ⇒ t = 0, 3.271658

tAmount of water in the pipe

0

3.271658

8

30

27.964561

48.543686

The amount of water in the pipe is a minimum at time t = 3.272 (or 3.271) hours.

(d) \(30+\int_{0}^{w}\left [ R(t)-D(t) \right ]dt.=50\)

Question

The figure above shows the graph of f’, the derivative of a twice-differentiable function f, on the closed interval 0 ≤ x ≤ 8. The graph of f’ has horizontal tangent lines at x = 1, x = 3, and x = 5. The areas of the regions between the graph of f’ and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f(8) = 4.
(a) Find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer.
(b) Determine the absolute minimum value of f on the closed interval 0 ≤ x ≤ 8. Justify your answer.
(c) On what open intervals contained in 0 < x < 8 is the graph of f both concave down and increasing? Explain your reasoning.
(d) The function g is defined by \(g(x)=(f(x))^{3}. If f(3)=-\frac{5}{2},\) find the slope of the line tangent to the graph of g at x = 3.

Answer/Explanation

Ans:

(a)  x = 6 is the only critical point at which f’ changes sign from negative to positive. Therefore, f has a local minimum at x = 6.
(b) From part (a), the absolute minimum occurs either at x = 6 or at an endpoint.

\(f(0)=f(8)+\int_{8}^{0}f'(x)dx\)

\(f(8)-\int_{8}^{0}f'(x)dx=4-12 = -8\)

\(f(6)=f(8)+\int_{8}^{6}f'(x)dx\)

\(=f(8)-\int_{8}^{6}f'(x)dx=4-7=-3\)

f(8) = 4

The absolute minimum value of f on the closed interval [0, 8] is -8.

(c) The graph of f is concave down and increasing on 0< x < 1 and 3 < x < 4,  because f’ is decreasing and positive on these intervals.

(d) \(g'(x)=3[f(x)]^{2}\cdot f'(x)\)

\(g'(3)=3[f(x)]^{2}\cdot f'(3)=3\left ( -\frac{5}{2} \right )^{2}\cdot 4=75\)

Question

The function f is defined on the closed interval [5, 4] . The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by \(g(x)=\int_{-3}^{x}f(t)dt.\)

(a) Find g(3).
(b) On what open intervals contained in -5 < x < 4 is the graph of g both increasing and concave down? Give a reason for your answer.
(c) The function h is defined by \(h(x)=\frac{g(x)}{5x}.\) Find h'(3).
(d) The function p is defined by \(p(x)=f(x^{2}-x).\) Find the slope of the line tangent to the graph of p at the point where x = -1.

Answer/Explanation

Ans:

(a) \(g(3)=\int_{-3}^{3}f(t)dt=6+4-1=9\)

(b) g′(x) = f ( x)
The graph of g is increasing and concave down on the intervals − 5< x < -3 and 0 < x < 2 because g′ = f is positive and decreasing on these intervals.

(c) \(h'(x)=\frac{5xg'(x)-g(x)5}{(5x)^{2}}=\frac{5xg'(x)-5g(x)}{25x^{2}}\)

\(h'(3)=\frac{(5)(3)g'(3)-5g(3)}{25.3^{2}}\)

\(=\frac{15(-2)-5(9)}{225}=\frac{-75}{225}=-\frac{1}{3}\)

(d) p'(x) = f'(x2 – x) (2x – 1)

p’ (-1) = f'(2) (-3) = (-2) (-3) = 6

Question

\(f(x)=1-\sqrt[3]{x}\)
(A) Find the intervals on which f is increasing or decreasing.
(B) Locate all maxima and minima.
(C) Find the intervals over which f is concave upward or downward.
(D) Find all inflection points.
(E) Sketch the graph of f.

Answer/Explanation

(A)Use the first derivative to find where f is increasing or decreasing. The first derivative \(f'(x)=\frac{1}{3}x^{-\frac{2}{3}}< 0\) for (-∞, +∞), then f is decreasing for all values of x.
(B) First, find any points where f ′(x) = 0, but f ′ < 0 for all values of x. At x = 0, f ′(x) does not exist since there is a zero in the denominator. Therefore, there are no maxima or minima.
(C) Use the second derivative to test for concavity.\)f”(x)=\frac{2}{9}x^{-\frac{5}{3}}\).f”>0  when  x > 0; therefore, f is concave upward on the interval (0, +∞). If x < 0, then f ″ < 0; therefore, f is concave downward on (-∞, 0).
(D) The second derivative f ″ changes concavity at x = 0, so there is an inflection point at x = 0. The function f (0) = 1, so the inflection point is at coordinate (0, 1), which is also the y-intercept.
(E) y-intercept at (0, 1). x-intercept at (1, 0).

Question

 Given the graph of f ′, find the following properties of the function f :

 

(A) The intervals on which f is increasing or decreasing
(B) The location of the relative maxima and minima
(C) The points of inflection and concavity of f
(D) Draw a sketch of f , given that f (-1) = f (1) = 5, f (0) = 0, and f (5) = -5.

Answer/Explanation

(A) f ′ < 0 on (-∞, -1) and (1, 5), so f is decreasing. f ′ > 0 on (-1, 1) and (5, +∞), so f is increasing.
(B) The function has a relative maximum at x = 1 since f ′ changes from positive to negative. There are relative minima at x = (-1, 5) since f ′ changes from negative to positive.

(C) f ′ is increasing on (-∞, 0), so f ′′ > 0 and f is concave upward. f ′ is decreasing on \(\left ( 0,2\frac{1}{2} \right )\) so f ′′ < 0 and f is concave downward. f ′ is increasing on\(\left ( 2\frac{1}{2} ,+\infty \right ) \), so f ′′ > 0 and f is concave upward. A change of concavity occurs at x = 0 and \(x=2\frac{1}{2}\) .Summarizing the results in a table:

Question

 \(f(x) =x^{4}-x^{3}\).
(A) Find the intervals on which f is increasing or decreasing.
(B) Locate all maxima and minima.
(C) Find the points of inflection, if any, on f .
(D) Find the intervals where f is concave upward or downward.
(E) Sketch the graph of \(f(x) =x^{4}-x^{3}\).

Answer/Explanation

(A) To figure out where f (x) is increasing or decreasing, set \(f'(x)=4x^{3}-3x^{2}=x^{2}(4x-3)=0\)
 The function equals zero at x = 0 and x = 3. Use a test point on each interval to find where f (x) is increasing or decreasing. The function f (x) is increasing on interval \(\left ( \frac{3}{4},\infty \right )\)
 but is decreasing for interval (-∞, 0) and decreasing for  \(\left ( 0,\frac{3}{4}\right )\).

B) You found in part (a) that f ′(x) = 0 at x = 0 and \(x=\frac{3}{4}\) . You then take the second derivative at those points to determine if they represent a maximum or minimum.You find that \(f{}”(x)=12x^{2}-6x\) You then substitute values for x to find that f ′′(0) = 0 and \(f”\left ( \frac{3}{4} \right )=\frac{9}{4}\). This demonstrates that \(f\left ( \frac{3}{4} \right )\) is a relative minimum. The test for f ′′(0) is inconclusive, so look back to the first derivative. The first derivative changes from positive to negative at x = 0, which implies that x = 0 is a relative maximum. At \(x=\frac{3}{4}\) ,f’=0 \(f”\left (\frac{3}{4}\right )=12\left ( \frac{9}{16} \right )-6\left ( \frac{3}{4} \right )=\frac{108}{16}-\frac{72}{16}=\frac{36}{16}> 0\) so \(\frac{3}{4}\) is a relative minimum.\(f\left ( \frac{3}{4} \right )=\left ( \frac{3}{4} \right )^{2}-\left ( \frac{3}{4} \right )^{4}-\left ( \frac{3}{4} \right )^{3}=\frac{81}{256}-\frac{108}{256}=-\frac{27}{256},\)so relative minimum at \(f\left ( \frac{3}{4} \right )=-\frac{27}{256}\).

 

Question

 Let \(g(t)=\int_{0}^{t}f(x)dx\) , and consider the graph of f shown below.

(A) Evaluate g(0), g(2), and g(6).
(B) On what interval(s) is g increasing (if any)? Justify your answer.
(C) At what value(s) of t does g have a minimum value? Justify your answer.
(D) On what interval(s) is g concave down? Justify your answer.

Answer/Explanation

(A)\(g(0)=\int_{0}^{0}f(x)dx=0\)
\(g(2)=\int_{0}^{2}f(x)dx=\int_{0}^{2}(4-4x)dx=\left [ 4x-2x^{2} \right ]^{2}_{0}=0\)
\(g(6)=\int_{0}^{6}f(x)dx=\int_{0}^{2}(4-4x)dx+\int_{3}^{2}(2x-8)dx+\int_{3}^{5}(4x-14)dx+\int_{5}^{6}6dx\)
\(=0+(-3)+4+6=7\)
(B) g is increasing on \([0,1)\cup (\frac{7}{2},6]\),since g'(t)=f(t)>0 on these intervals. Note that \(\frac{7}{2}\) is the solution to 4x -14 = 0.
(C) At t=7/2,g has a minimum value. g(0)=0, g(7/2)=-7/2, g(6)=7
(D) Since g′(t) = f(t) is decreasing only on (0, 2), you see that g′′(x)<0 on this interval. Therefore, g is concave down only on (0, 2).

Question

x11.11.21.31.4
f'(x)810121314.5

The function f is twice differentiable for x > 0 with f(1) = 15 and f”(1) = 20. Values of f’, the derivative of f, are given for selected values of x in the table above.
(a) Write an equation for the line tangent to the graph of f at x =1. Use this line to approximate f(1.4).
(b) Use a midpoint Riemann sum with two subintervals of equal length and values from the table to approximate \(\) Use the approximation for \(\) to estimate the value of f(1.4). Show the computations that lead to your answer.
(c) Use Euler’s method, starting at x = 1 with two steps of equal size, to approximate f(1.4). Show the computations that lead to your answer.
(d) Write the second-degree Taylor polynomial for f about x = 1. Use the Taylor polynomial to approximate f(1.4).

Answer/Explanation

Ans:

(a) f(1) = 15, f'(1) = 8

An equation for the tangent line is
y  = 15 + 8 (x – 1) .

f(1.4) ≈ 15 + 8 (1.4 – 1) = 18.2

(b) \(\int_{1}^{1.4}f'(x)dx\approx (0.2)(10)+(0.2)(13)=4.6\)

\(f(1.4)=f(1)+\int_{1}^{1.4}f'(x)dx\)

f(1.4) ≈ 15 + 4.6 = 19.6

(c) f(1.2) ≈ f(1) + (0.2) (8) = 16.6

f(1.4) ≈ 16.6 + (0.2) (12) = 19.0

(d) \(T_{2}(x)=15+8(x-1)+\frac{20}{2!}(x-1)^{2}\)

= 15 + 8 (x-1) + 10(x – 1)2

f(1.4) ≈ 15 + 8(1.4-1) + 10(1.4 – 1)2 = 19.8

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