Question
Let f be a continuous function defined on the closed interval −4 ≤ x ≤ 6. The graph of f , consisting of four line segments, is shown above. Let G be the function defined by \(G(x)=\int_{0}^{x}f(t)dt.\)
(a) On what open intervals is the graph of G concave up? Give a reason for your answer.
(b) Let P be the function defined by P (x) = G (x)• f(x). Find P'(3 ).
(c) Find \(\lim_{x\rightarrow 2}\frac{G(x)}{x^{2}-2x}.\)
(d) Find the average rate of change of G on the interval [−4, 2]. Does the Mean Value Theorem guarantee a value c, −4 < c < 2, for which G'(c) is equal to this average rate of change? Justify your answer.
Answer/Explanation
Ans:
(a)
G'(x) = f(x)
G”(x) = f'(x)
On (-4, -2) and (2, 6), G(x) is concave up because f(x) (which is equal to G'(x)) has a positive slope / is increasing.
(b)
p'(x) = G'(x) f(x) + f'(x) G(x)
p'(3) = G'(3) f(3) + f'(3) G(3)
G'(x) = f(x) \(G(3)=\int_{0}^{3}f(t)dt=-\frac{7}{2}\)
\(p'(3)=(-3)(-3)+(1)\left (- \frac{7}{2} \right )\)
(c)
\(\lim_{x\rightarrow 2}G(x)=\lim_{x\rightarrow 2}(x^{2}-2x)=0\) Must be I’hopital’s rule
\(\int_{0}^{2}f(t)dt=0\)
\(\lim_{x\rightarrow 2}\frac{G'(x)}{2x-2}=\frac{f(2)}{4-2}=\frac{-4}{2}\)
(d)
\(\int_{0}^{2}f(t)dt\) \(\int_{0}^{-4}f(t)dt= -(3+9+3+1)\)
= -16
AROC of \(G = \frac{G(2)-G(-4)}{2-(-4)}=\frac{0-(16)}{2+4}=\frac{16}{6}=\frac{8}{3}\)
The meanvalue theorem does guarantee a value of -4<c<2, for which G'(c) is equal to average rate of chang. This is because G'(x) = f(t) and x = t exists for all values, -4<x=+t<2, meaning that G(x) is continuous on the closed interval and differentiable on the open interval.
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
\(f(x)=1-\sqrt[3]{x}\)
(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).