The derivative of the function $f$ is defined by $f^{\prime}(x)=x^{2}-3 x+3 \sin (2 x+1)$ . Find the $x$ values, if any, in the interval $-2.5<x<2.5$ 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}-3 x+3 \sin (2 x+1)

. Find the

x

values, if any, in the interval

-2.5<x<2.5

where the function

f

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

3

decimal places.

\newline

Answer:

x=

Find Critical Points: To find the relative minimum of the function $f$ , we need to find the critical points of $f$ by setting its derivative $f'(x)$ equal to zero and solving for $x$ . The derivative is given by $f'(x) = x^2 - 3x + 3\sin(2x+1)$ .
Set Derivative Equal: Set the derivative equal to zero to find the critical points: $x^2 - 3x + 3\sin(2x+1) = 0$ . This equation cannot be solved algebraically due to the presence of the sine function, so we will use a calculator to find the solutions within the interval $-2.5 < x < 2.5$ .
Graph Function: Using a calculator, we graph the function $y = x^2 - 3x + 3\sin(2x+1)$ and look for the $x$ -values where the graph crosses the $x$ -axis within the interval $-2.5 < x < 2.5$ . These $x$ -values are the critical points where $f$ could have a relative minimum.
Find $X$ -Values: After graphing, we find that the function crosses the $x$ -axis at certain points. We use the calculator's feature to find the exact $x$ -values to three decimal places.
Test Critical Points: Suppose the calculator gives us one or more $x$ -values where the function crosses the $x$ -axis. We then need to test these $x$ -values to determine if they correspond to relative minima. We do this by using the first or second derivative test.
First Derivative Test: For the first derivative test, we look at the sign of $f'(x)$ just 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.
Second Derivative Test: For the second derivative test, we would find $f''(x)$ and evaluate it at the critical points. If $f''(x)$ is positive at a critical point, then $f$ has a relative minimum at that point. However, since the second derivative involves the cosine function and is more complex, we will use the first derivative test.
Apply First Derivative Test: After applying the first derivative test to the critical points found, we determine which of them are relative minima. We list these $x$ -values as the answer.