Find the absolute extreme values of the function on the interval.
-$$f ( x ) = e ^ { x } - x , - 4 \leq x \leq 2$$
A) Minimum value is $\mathrm { e } ^ { - 4 } + 4$ at $\mathrm { x } = - 4$; maximum value is $\mathrm { e } ^ { 2 } - 2$ at $\mathrm { x } = 2$
B) Minimum value is 1 at $x = 0$; no maximum value
C) Minimum value is 1 at $x = 0$; maximum value is $\mathrm { e } ^ { 2 } - 2$ at $x = 2$
D) Minimum value is 1 at $x = 0$; maximum value is $e ^ { - 4 } + 4$ at $x = - 4$

Question

Quizplus · Accepted Answer

To find the extreme values of the function, we need to first take the derivative and set it equal to zero to find the critical points.

$f'(x) = e^x - 1$

Setting it equal to zero, we get:

$e^x - 1 = 0$

$e^x = 1$

$x = 0$ is a critical point.

We also need to check the endpoints of the interval, which are $x = -4$ and $x = 2$.

Plugging in these values to the function, we get:

$f(-4) = e^{-4} + 4 \approx 4.018$

$f(0) = e^0 - 0 = 1$

$f(2) = e^2 - 2 \approx 5.389$

Thus, we have a minimum value of $1$ at $x = 0$, and a maximum value of $e^2 - 2$ at $x = 2$.

Therefore, the best choice is C.

Find the Absolute Extreme Values of the Function on the Interval