Question

the derivative of the function f is given by f′(x)=e−xcos(x2), for all real numbers x. what is the minimum value of f(x) for −1≤x≤1?

asked Feb 26, 2024 179k views

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Mohamida · Answer 1 · 2024-03-02T21:16:06+0000

There is no minimum value of
f(x) for
-1 < = x < = 1 . Therefore, The correct option is d.

To find the minimum value of
$f(x)$ within the given interval
$-1 \leq x \leq 1$ , we need to analyze the critical points and endpoints of the interval.

Given
$f'(x) = e^(-x) \cdot \cos(x^2)$ , we first need to find the critical points by setting the derivative equal to zero:

$f'(x) = 0$

$e^(-x) \cdot \cos(x^2) = 0$

This equation is satisfied when either
$e^(-x) = 0$ (which is impossible) or when
$$\cos(x^2) = 0$.$

For
$\cos(x^2) = 0\), \(x^2$ must be of the form
$(\pi)/(2) + n\pi$ where
$n$ is an integer.

This implies that
$x$ can take values such that
$x^2 = (\pi)/(2), (3\pi)/(2), (5\pi)/(2), \ldots$ etc. within the interval
$$-1 \leq x \leq 1$.$

However,
$x^2 = (\pi)/(2)$ does not satisfy the interval condition, so the only critical point in the interval
$-1 \leq x \leq 1$ occurs when
$x^2 = (3\pi)/(2)$ , yielding
$x = \sqrt{(3\pi)/(2)}$ or
$$x = -\sqrt{(3\pi)/(2)}$.$

But
$\sqrt{(3\pi)/(2)}$ exceeds the limit of
$x \leq 1$ in the interval, so the only critical point within the interval is
$$x = -\sqrt{(3\pi)/(2)}$.$

Now we need to evaluate
$f(x)$ at the endpoints
$($-1$)$ and
$$(1$)$ and the critical point within the interval.

$f(-1), f(1), \text{ and } f\left(-\sqrt{(3\pi)/(2)}\right)$

$f(-1) = e^(-(-1)) \cdot \cos((-1)^2) = e \cdot \cos(1)$

$f(1) = e^(-1) \cdot \cos(1)$

$f\left(-\sqrt{(3\pi)/(2)}\right) = e^{\sqrt{(3\pi)/(2)}} \cdot \cos\left(-(3\pi)/(2)\right) = e^{\sqrt{(3\pi)/(2)}} \cdot 0 = 0$

Among these values, the minimum value is
$$f\left(-\sqrt{(3\pi)/(2)}\right) = 0$.$

Therefore, the correct answer is No minimum value of
$$f(x)$ for $-1 \leq x \leq 1$.$