Question
| t (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| C(t) (ounces) | 0 | 5.3 | 8.8 | 11.2 | 12.8 | 13.8 | 14.5 |
Hot water is dripping through a coffeemaker, filling a large cup with coffee. The amount of coffee in the cup at time t, 0 ≤ t ≤ 6, is given by a differentiable function C, where t is measured in minutes. Selected values of C(t), measured in ounces, are given in the table above.
(a) Use the data in the table to approximate C'(3.5). Show the computations that lead to your answer, and indicate units of measure.
(b) Is there a time t, 2 ≤ t ≤ 4 at which C'(t) = 2 ? Justify your answer.
(c) Use a midpoint sum with three subintervals of equal length indicated by the data in the table to approximate the value of \(\frac{1}{6}\int_{0}^{6}C(t)dt.\) Using correct units, explain the meaning of \(\frac{1}{6}\int_{0}^{6}C(t)dt\) in the context of the problem.
(d) The amount of coffee in the cup, in ounces, is modeled by B(t) = 16 – 16e-0.4t. Using this model, find the rate at which the amount of coffee in the cup is changing when t = 5.
▶️Answer/Explanation
Ans:
(a) \(C'(3.5)\approx \frac{C(4)-C(3)}{4-3}=\frac{12.8-11.2}{1}=1.6 ounces/min\)
(b) C is differentiable ⇒ C is continuous (on the closed interval) \(\frac{C(4)-C(2)}{4-2}=\frac{12.8-8.8}{2}=2\)
Therefore, by the Mean Value Theorem, there is at least
one time t, 2 < t < 4, for which C'(t)=2.
(c) \(\frac{1}{6}\int_{0}^{6}C(t)dt\approx \frac{1}{6}\left [ 2\cdot C(1)+2\cdot C(3)+2\cdot C(5) \right ]\)
\(= \frac{1}{6}(2\cdot 5.3+2\cdot 11.2+2\cdot 13.8)\)
\(= \frac{1}{6}(60.6)= 10.1 ounces\)
\(\frac{1}{6}\int_{0}^{6}C(t)dt\) is the average amount of coffee in the cup, in ounces, over the time interval 0 ≤ t ≤ 6 minutes.
(d) \(B'(t)=-16(-0.4)e^{-0.4t}=6.4e^{-0.4t}\)
\(B'(5)=6.4e^{-0.4(5)}=\frac{6.4}{e^{2}}ounces/min\)
Question:
Let f be a function defined by \(\left\{\begin{matrix}
1-2 sin x & for x \leq 0 & & \\e^{-4x}
&for x > 0. & &
\end{matrix}\right.\)
(a) Show that f is continuous at x = 0.
(b) For x ≠ 0, express f'(x) as a piecewise-defined function. Find the value of x for which f'(x) = -3.
(c) Find the average value of f on the interval [-1, 1] .
▶️Answer/Explanation
Ans:
(a)
To be continuous
i) f(0) = 1
ii) \(\lim_{x\rightarrow 0} 1 -2 sin x = \lim_{x\rightarrow 0}e^{-4x}\)
1 = 1 ∴ \(\lim_{x\rightarrow 0^{-}} f = \lim_{x\rightarrow 0^{+}}f\)
∴ \(\lim_{x\rightarrow 0} f = 1\)
iii) f(0) = \(\lim_{x\rightarrow 0} f = 1\)
∴ f is continuous for all values of x.
(b)
\(f'(x)=\left\{\begin{matrix}
-2 cos x &, &x<0 & \\-4e^{-4x}
&, &x>0 &
\end{matrix}\right.\)
\({f}'(x) = 3\)
Since -2 cos x oscillates between -2 and 2 there will be no such value in this function such that f'(x) = -3
but \({f}'(x) =- 3\)
-4e -4x = -3
e -4x = 3/4
-4x = In (3/4) \(\therefore = \frac{-1}{4}In\left ( \frac{3}{4} \right )\therefore f’\left ( \frac{-1}{4}In\left ( \frac{3}{4} \right ) \right )= -3\)
(c)
\(f_{avg} = \frac{1}{1+1}\int_{-1}^{1}f(x)dx\)
\(= \frac{1}{2}\left [ \int_{-1}^{0} 1-2 sinx dx + \int_{0}^{1}e^{-4x}dx\right ]\)
\(= \frac{1}{2}\left [ [x+2cosx]_{-1}^{0}+[\frac{-1}{4}e^{-4x}]_{0}^{1} \right ]\)
\(= \frac{1}{2}\left [ 2+1-2cos(-1)\frac{-e^{-4}}{4} +\frac{1}{4}\right ]\)
\(=\frac{-\left [ 3-2cos(-1) \right ]+\left [ \frac{-1}{4e^{4}} +\frac{1}{4}\right ]}{2}\)
Question:
The twice-differentiable functions f and g are defined for all real numbers x. Values of f, f’, g, and g’ for various values of x are given in the table above.
(a) Find the x-coordinate of each relative minimum of f on the interval [-2, 3] . Justify your answers.
(b) Explain why there must be a value c, for -1 < c < 1, such that f”(c) =0.
(c) The function h is defined by h(x) = In (f(x)). Find h'(3). Show the computations that lead to your answer.
(d) Evaluate \(\int_{-2}^{3}f'(g(x))g'(x)dx.\)
▶️Answer/Explanation
Ans:
(a)
Critical numbers : x = -1, 1
f has a rel. min. at x = 1 because f'(1) = 0 and f’ switches sign from negative to positive there.
(b)
f'(-1) = 0 and f'(1) = 0, and f'(x) is differentiable and continuous on the internal so by rolle’s Theorem there is some value c where f”(c) = 0.
(c)
\(h'(x)=\frac{f'(x)}{f(x)}\)
\(h'(3)=\frac{f'(3)}{f(3)}\)
\(h'(3)=\frac{\frac{1}{0}}{7}\)
\(h'(3)=\frac{1}{2}\cdot \frac{1}{7}\)
\(h'(3)=\frac{1}{14}\)
(d)
u = g(x)
du = g'(x) dx
\(\int f'(u)du\)
\(\left [ f\left ( g(x) \right ) \right ]_{-2}^{3}\)
\(f(g(3)) – f(g(-2))\)
\(f(1) – f(-1)\)
\(2-8\)
\(-6\)
Question:
The function f is differentiable on the closed interval [−6, 5 ] and satisfies f (−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)
\(f(-6)=\left ( \int_{-2}^{-6}f'(x)dx \right )+f(-2)\)
f(-6) = 3
\(f(5)=f(-2)+\int_{-2}^{5}f'(x)dx\)
f(5) = 10 – 2π
(b)
f is increasing on x = [-6, -2]
u [2, 5], since f’> 0 on the interval x ∈ [-6, -2] v[2, 5]
(c)
The absolute minimum endpoints of f or [-6, 5] is f(-6) = 3
7 – 2 π , since f(5) = 10 – 2π
f(2) < f (5) and f(-6) critical points
the endpoints f’ = 0
and f(2) < f(-2) the f(-2) = 7
other critical points, f(2) = 7.2 π
by EVT
(d)
\(f”(-5)=\frac{-1}{2}\)
f”(3) = DNE, as the
\(\lim_{x\rightarrow 3^{+}}\frac{f'(x)-2}{x-3}\neq \lim_{x\rightarrow 3^{-}}\frac{f'(x)-2}{x-3}\)
Therefore it is impossible to take a derivative at x = 3 in f’