Answer
(a) Using the graph, we can make the following estimates:
There is no absolute maximum value, but the local maximum is approximately $f(2) \approx 0.5$
The minimum value appears to be $f(0) = 0$
The exact values are as follows:
The local maximum value is $f(2) = \frac{4}{e^2} \approx 0.54$
The minimum value is $f(0) = 0$
(b) The graph appears to increase most rapidly, that is it has the highest positive slope, approximately at the point $x = 0.5$
The exact value is as follows:
The slope of the graph is a maximum when $x = 2 - \sqrt{2} \approx 0.586$
Work Step by Step
(a) $f(x) = x^2~e^{-x}$
Using the graph, we can make the following estimates:
There is no absolute maximum value, but the local maximum is approximately $f(2) \approx 0.5$
The minimum value appears to be $f(0) = 0$
We can find the exact values:
$f'(x) = 2x~e^{-x}-x^2~e^{-x}$
$f'(x) = e^{-x}(2x-x^2) = 0$
$2x-x^2 = 0$
$x(x-2) = 0$
$x = 0, 2$
$f(0) = 0^2~e^{-0} = 0$
$f(2) = 2^2~e^{-2} = \frac{4}{e^2} \approx 0.54$
The local maximum value is $f(2) = \frac{4}{e^2} \approx 0.54$
The minimum value is $f(0) = 0$
(b) The graph appears to increase most rapidly, that is it has the highest positive slope, approximately at the point $x = 0.5$
We can find $x$ when $f''(x) = 0$:
$f''(x) = 2~e^{-x}-2x~e^{-x}-2x~e^{-x}+x^2~e^{-x}$
$f''(x) = x^2~e^{-x}-4x~e^{-x}+2~e^{-x}$
$f''(x) = (x^2-4x+2)~e^{-x} = 0$
$x^2-4x+2 = 0$
We can use the quadratic formula:
$x = \frac{4\pm \sqrt{(-4)^2-(4)(1)(2)}}{2(1)}$
$x = \frac{4\pm \sqrt{16-8}}{2}$
$x = \frac{4\pm 2\sqrt{2}}{2}$
$x = 2 \pm \sqrt{2}$
Using the graph, we can clearly see that the negative branch is the correct solution.
The slope of the graph is a maximum when $x = 2 - \sqrt{2} \approx 0.586$