The derivative of the function $f$ is defined by $f^{\prime}(x)=x^{2}+2 x-3 \sin (3 x)$ . Find the $x$ values, if any, in the interval $-2.5<x<3$ where the function $f$ has a relative minimum. You may use a calculator and round all values to $3$ decimal places. $\newline$ Answer: $x=$

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Q. The derivative of the function

f

is defined by

f^{\prime}(x)=x^{2}+2 x-3 \sin (3 x)

. Find the

x

values, if any, in the interval

-2.5<x<3

where the function

f

has a relative minimum. You may use a calculator and round all values to

3

decimal places.

\newline

Answer:

x=

Identify Critical Points: Identify the critical points of the function $f$ by setting its derivative $f'(x)$ equal to zero and solving for $x$ . $f'(x) = x^2 + 2x - 3\sin(3x) = 0$ This is a transcendental equation and may not have an algebraic solution. We will use a calculator to find the roots in the interval $-2.5 < x < 3$ .
Find Zero of Derivative: Use a graphing calculator or numerical methods to find the approximate values of $x$ where $f'(x) = 0$ in the given interval. $\newline$ By graphing or using a numerical solver, we can find the $x$ values that make the derivative zero. These $x$ values are the potential points where $f$ could have a relative minimum or maximum.
Determine Relative Minimum: Determine whether each critical point is a relative minimum by using the second derivative test or the first derivative test. $\newline$ For the second derivative test, we would find $f''(x)$ and evaluate it at the critical points. If $f''(x) > 0$ , then $f$ has a relative minimum at that point. If $f''(x) < 0$ , then $f$ has a relative maximum at that point. $\newline$ Alternatively, for the first derivative test, we would look at the sign of $f'(x)$ before and after each critical point. If $f'(x)$ changes from negative to positive at a critical point, then $f$ has a relative minimum at that point.
Calculate Second Derivative: Use the calculator to find the second derivative of $f$ , if necessary, and evaluate it at the critical points found in Step $2$ . $\newline$ Since the second derivative of $f$ involves the derivative of $-3\sin(3x)$ , which is $-9\cos(3x)$ , the second derivative $f''(x)$ will be $f''(x) = 2x + 2 - 9\cos(3x)$ . $\newline$ We can then use the calculator to evaluate $f''(x)$ at each critical point to determine if it is a relative minimum.
Report Minimum Values: Report the $x$ values where $f$ has a relative minimum, rounding to three decimal places as instructed. $\newline$ After evaluating the second derivative at the critical points or using the first derivative test, we can identify the $x$ values where $f$ has a relative minimum. These values should be within the interval $-2.5 < x < 3$ .