Let $f(x)=-4 x^{3}+6 x^{2}+1$ . $\newline$ What is the absolute minimum value of $f$ over the closed interval $-4 \leq x \leq 3$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $3$ $\newline$ (B) $-78$ $\newline$ (C) $-353$ $\newline$ (D) $-53$

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Euler's method

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Q. Let

f(x)=-4 x^{3}+6 x^{2}+1

\newline

What is the absolute minimum value of

f

over the closed interval

-4 \leq x \leq 3

\newline

Choose

1

answer:

\newline

(A)

3

\newline

(B)

-78

\newline

(C)

-353

\newline

(D)

-53

Find Derivative: To find the absolute minimum value of the function on the closed interval, we need to find the critical points of the function within the interval and evaluate the function at the endpoints of the interval.
Locate Critical Points: First, we find the derivative of the function $f(x)$ to locate the critical points. The derivative $f'(x)$ is given by: $\newline$ $f'(x) = \frac{d}{dx} (-4x^{3} + 6x^{2} + 1) = -12x^{2} + 12x$ .
Evaluate Function: Next, we set the derivative equal to zero to find the critical points: $\newline$ $-12x^{2} + 12x = 0$ $\newline$ $x(12x - 12) = 0$ $\newline$ $x = 0$ or $x = 1$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ .Evaluating $f(x)$ at $x = -4$ : $f(-4) = -4(-4)^{3} + 6(-4)^{2} + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ .Evaluating $f(x)$ at $x = -4$ : $f(-4) = -4(-4)^{3} + 6(-4)^{2} + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353$ .Evaluating $f(x)$ at $x = 0$ : $f(-4)$ $0$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ .Evaluating $f(x)$ at $x = -4$ : $f(-4) = -4(-4)^{3} + 6(-4)^{2} + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353$ .Evaluating $f(x)$ at $x = 0$ : $f(-4)$ $0$ .Evaluating $f(x)$ at $f(-4)$ $2$ : $f(-4)$ $3$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ .Evaluating $f(x)$ at $x = -4$ : $f(-4) = -4(-4)^{3} + 6(-4)^{2} + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353$ .Evaluating $f(x)$ at $x = 0$ : $f(-4)$ $0$ .Evaluating $f(x)$ at $f(-4)$ $2$ : $f(-4)$ $3$ .Evaluating $f(x)$ at $f(-4)$ $5$ : $f(-4)$ $6$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ .Evaluating $f(x)$ at $x = -4$ : $f(-4) = -4(-4)^{3} + 6(-4)^{2} + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353$ .Evaluating $f(x)$ at $x = 0$ : $f(-4)$ $0$ .Evaluating $f(x)$ at $f(-4)$ $2$ : $f(-4)$ $3$ .Evaluating $f(x)$ at $f(-4)$ $5$ : $f(-4)$ $6$ .Comparing the values of $f(x)$ at the critical points and endpoints, we find that the smallest value is $f(-4)$ $8$ . However, since we are looking for the absolute minimum, we need to consider the negative of this value, which is $f(-4)$ $9$ .
Calculate Absolute Minimum: Now we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval: $f(-4)$ , $f(0)$ , $f(1)$ , and $f(3)$ . Evaluating $f(x)$ at $x = -4$ : $f(-4) = -4(-4)^{3} + 6(-4)^{2} + 1 = -4(-64) + 6(16) + 1 = 256 + 96 + 1 = 353$ . Evaluating $f(x)$ at $x = 0$ : $f(-4)$ $0$ . Evaluating $f(x)$ at $f(-4)$ $2$ : $f(-4)$ $3$ . Evaluating $f(x)$ at $f(-4)$ $5$ : $f(-4)$ $6$ . Comparing the values of $f(x)$ at the critical points and endpoints, we find that the smallest value is $f(-4)$ $8$ . However, since we are looking for the absolute minimum, we need to consider the negative of this value, which is $f(-4)$ $9$ . The absolute minimum value of the function $f(x)$ over the closed interval $f(0)$ $1$ is $f(-4)$ $9$ .