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

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The derivative of the function  f is defined by  f^(')(x)=x^(2)-3x+3sin(2x+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. Answer:  x=

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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 + 3sin(2x+1).Set Derivative Equal: Set the derivative equal to zero to find the critical points: x^2 - 3x + 3sin(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 + 3sin(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.

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

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The derivative of the function f f f is defined by f′(x)=x2−3x+3sin⁡(2x+1) f^{\prime}(x)=x^{2}-3 x+3 \sin (2 x+1) f′(x)=x2−3x+3sin(2x+1). Find the x x x values, if any, in the interval \( -2.5

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