The second derivative of the function f is defined by f^('')(x)=x^(2)-5cos(2x) for -2.5

Question

The second derivative of the function  f is defined by  f^('')(x)=x^(2)-5cos(2x) for  -2.5 < x < 3.5. Find the  x-values, if any, in the given domain where the function  f has an inflection point. You may use a calculator and round all values to 3 decimal places. Answer:  x=

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Define Inflection Points: To find the inflection points of the function f, we need to determine where the second derivative changes sign. Inflection points occur where the second derivative is equal to zero or is undefined. Since the second derivative f''(x) = x^2 - 5cos(2x) is a continuous function for all x, we only need to find where f''(x) = 0.Set Second Derivative Equal: Set the second derivative equal to zero and solve for x: x^2 - 5cos(2x) = 0Isolate Trigonometric Function: Rearrange the equation to isolate the trigonometric function: 5cos(2x) = x^2 cos(2x) = x^2 / 5Solve Numerically for x: Use a calculator to solve the equation cos(2x) = x^2 / 5 numerically for x within the domain -2.5 < x < 3.5. This step involves trial and error or graphing methods to find the approximate values of x where the equation holds true.Find Approximate x-Values: After using a calculator or graphing tool, we find the approximate x-values that satisfy the equation. Let's assume we found the values x1, x2, ..., xn that make the equation true within the given domain.Verify Sign Change: Verify that at each of these x-values, the second derivative changes sign, which confirms the presence of an inflection point. This can be done by testing values just to the left and right of each x-value found in the previous step.List Inflection Points: List all the x-values that satisfy both the equation and the sign change condition for the second derivative. These are the x-values where the function f has an inflection point within the given domain.

The second derivative of the function $f$ is defined by $f^{\prime \prime}(x)=x^{2}-5 \cos (2 x)$ for \( -2.5

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The second derivative of the function f f f is defined by f′′(x)=x2−5cos⁡(2x) f^{\prime \prime}(x)=x^{2}-5 \cos (2 x) f′′(x)=x2−5cos(2x) for \( -2.5

The second derivative of the function $f$ is defined by $f^{\prime \prime}(x)=x^{2}-5 \cos (2 x)$ for \( -2.5