Activity 44.
Find the exact absolute maximum and minimum of each function on the stated interval.
(a)
\(h(x) = xe^{-x}\text{,}\) \([0,3]\)
Hint.
After computing \(h'(x)\text{,}\) factor to write the derivative as a product.
Answer.
Absolute maximum: \(e^{-1}\text{;}\) absolute minimum: \(0\text{.}\)
Solution.
For \(h(x) = xe^{-x}\text{,}\) we know that \(h'(x) = xe^{-x}(-1) + e^{-x} = e^{-x}(-x+1)\text{.}\) Therefore, the only critical number of \(h\) is \(x = 1\text{.}\) Next, we compute \(h(1)\text{,}\) \(h(0)\text{,}\) and \(h(3)\text{.}\) Observe that
-
\(\displaystyle h(1) = e^{-1} \approx 0.36788\)
-
\(\displaystyle h(0) = 0\)
-
\(\displaystyle h(3) = 3e^{-3} \approx 0.14936\)
Thus, on \([0,3]\text{,}\) the absolute maximum of \(h\) is \(e^{-1}\) and the absolute minimum is \(0\text{.}\)
(b)
\(p(t) = \sin(t) + \cos(t)\text{,}\) \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
Hint.
The sine and cosine functions have the same value at \(\frac{\pi}{4} \pm k\pi\) for any integer \(k\text{.}\)
Answer.
Absolute maximum: \(\sqrt{2}\text{;}\) absolute minimum: \(-1\text{.}\)
Solution.
Given \(p(t) = \sin(t) + \cos(t)\text{,}\) it follows \(p'(t) = \cos(t) - \sin(t)\text{,}\) so \(p'(t) = 0\) implies that \(\cos(t) =\sin(t)\text{.}\) The sine and cosine functions have the same value at \(\frac{\pi}{4} \pm k\pi\) for any integer \(k\text{.}\) The only time this occurs in \([-\frac{\pi}{2}, \frac{\pi}{2}]\) is for \(x = \frac{\pi}{4}\text{,}\) and thus this is the only critical number of \(p\) in the given interval. Now,
-
\(\displaystyle p(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.41421\)
-
\(\displaystyle p(-\frac{\pi}{2}) = \sin(-\frac{\pi}{2}) + \cos(-\frac{\pi}{2}) = -1 + 0 = -1\)
-
\(\displaystyle p(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1\)
Therefore, on \([-\frac{\pi}{2},\frac{\pi}{2}]\text{,}\) the absolute maximum of \(p\) is \(\sqrt{2}\) and the absolute minimum is \(-1\text{.}\)
(c)
\(q(x) = \frac{x^2}{x-2}\text{,}\) \([3,7]\)
Hint.
Upon finding \(q'(x)\text{,}\) factor its numerator.
Answer.
Absolute maximum: 9.8; absolute minimum: 8.
Solution.
With \(q(x) = \frac{x^2}{x-2}\text{,}\) we have
\begin{equation*}
q'(x) = \frac{(x-2)(2x) - x^2(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2-4x}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}\text{.}
\end{equation*}
Hence, the critical numbers of \(q\) are \(x = 0\) and \(x = 4\text{.}\) Only the latter critical number lies in the interval \([3,7]\text{,}\) and thus we evaluate \(q\) and find
-
\(\displaystyle q(4) = \frac{16}{2} = 8\)
-
\(\displaystyle q(3) = \frac{9}{1} = 9\)
-
\(\displaystyle q(7) = \frac{49}{5} = 9.8\)
We now see that on \([3,7]\) the absolute maximum of \(q\) is 9.8 and the absolute minimum is 8.
(d)
\(f(x) = 4 - e^{-(x-2)^2}\text{,}\) \((-\infty, \infty)\)
Hint.
Remember that \(e^{-(x-2)^2}\) is never zero.
Answer.
Absolute minimum 3; no absolute maximum.
Solution.
Here, we first observe that we are working on the domain of all real numbers, not a closed bounded interval. Hence, we need to think about the overall behavior of the function. First, since \(f(x) = 4 - e^{-(x-2)^2}\text{,}\) by the chain rule we see that \(f'(x) = -e^{-(x-2)^2}(-2(x-2)) = 2(x-2)e^{-(x-2)^2}\text{.}\) Since \(e^{-(x-2)^2}\) is always positive (in particular, never zero), it follows that the only critical number of \(f\) is \(x = 2\text{.}\) Furthermore, with \(f'(x) = 2(x-2)e^{-(x-2)^2}\text{,}\) we see that for \(x \lt 2\text{,}\) \(f'(x) \lt 0\text{,}\) while for \(x \gt 2\text{,}\) \(f'(x) \gt 0\text{.}\) This tells us by the first derivative test that \(f\) is decreasing for \(x \lt 2\) and increasing for \(x \gt 2\text{,}\) which tells us that \(f\) has an absolute minimum at \(x = 2\text{,}\) and \(f\) does not have an absolute maximum.
(e)
\(h(x) = xe^{-ax}\text{,}\) \([0, \frac{2}{a}]\) (\(a \gt 0\))
Hint.
After differentiating, remove a factor of \(e^{-ax}\text{.}\)
Answer.
Absolute minimum \(0\text{;}\) absolute maximum \(\frac{1}{a}e^{-1}\text{.}\)
Solution.
For \(h(x) = xe^{-ax}\) on \([0, \frac{2}{a}]\text{,}\) where \(a \gt 0\text{,}\) we start by finding \(h'(x)\text{.}\) By the product and chain rules, treating \(a\) as a constant, we find
\begin{equation*}
h'(x) = xe^{-ax}(-a) + e^{-ax}\text{.}
\end{equation*}
Factoring, it follows
\begin{equation*}
h'(x) = e^{-ax}(-ax+1)\text{.}
\end{equation*}
Since \(e^{-ax}\) is never zero, the only way for \(h'(x)=0\) is if \(-ax + 1 = 0\text{,}\) which implies \(x = \frac{1}{a}\text{.}\) Noting also that \(h'(x)\) is defined for every real number \(x\text{,}\) we have established that \(h\) has a single critical number at \(x = \frac{1}{a}\text{.}\)
Now we simply compute the value of \(h\) at each endpoint and at the critical number and compare the outputs. We have \(h(0) = 0\text{,}\) \(h(\frac{1}{a}) = \frac{1}{a} e^{-1}\text{,}\) and \(h(\frac{2}{a}) = \frac{2}{a} e^{-2}\text{.}\) The absolute minimum is \(0\text{,}\) and the absolute maximum is the larger of \(\frac{1}{a} e^{-1}\) and \(\frac{2}{a} e^{-2}\text{.}\) That absolute maximum is \(\frac{1}{a} e^{-1}\text{,}\) since
\begin{equation*}
\frac{2}{a} e^{-2} \lt \frac{2}{a} \cdot \frac{1}{e} \frac{1}{2} = \frac{1}{a}e^{-1}\text{.}
\end{equation*}
(f)
\(f(x) = b - e^{-(x-a)^2}\text{,}\) \((-\infty, \infty)\text{,}\) \(a, b \gt 0\)
Hint.
Compare part (d).
Answer.
Absolute minimum \(b-1\text{;}\) no absolute maximum.
Solution.
The reasoning here is identical to that in (d), but instead of working with \(f(x) = 4 - e^{-(x-2)^2}\text{,}\) we use arbitrary positive constants \(a\) and \(b\) to consider \(f(x) = b - e^{-(x-a)^2}\)
