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The derivative of the function is defined by for \( 0
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Q. The derivative of the function is defined by for . Find the -values, if any, in the given domain where the function has an inflection point. You may use a calculator and round all values to decimal places.Answer:
- Find Second Derivative: To find the inflection points of the function , we need to find the second derivative of , denoted as , and then determine where this second derivative changes sign. The second derivative will give us the concavity of the function, and inflection points occur where the concavity changes from concave up to concave down or vice versa.
- Solve for Potential Inflection Points: First, we find the second derivative of . The derivative of is , and the derivative of with respect to is multiplied by the derivative of the inside function, which is . Therefore, .
- Use Calculator to Find -Values: Next, we need to find the -values where , because these are the potential inflection points. We set the second derivative equal to zero and solve for : .
- Check Sign Change for Inflection Points: We rearrange the equation to isolate the trigonometric function: .
- Check Sign Change for Inflection Points: We rearrange the equation to isolate the trigonometric function: . Now, we divide both sides by to get .
- Check Sign Change for Inflection Points: We rearrange the equation to isolate the trigonometric function: . Now, we divide both sides by to get . This equation involves both trigonometric and algebraic functions, and it's not straightforward to solve algebraically. We will use a calculator to find the -values that satisfy the equation within the given domain 0 < x < 3.5. We are looking for points where the graph of intersects with the graph of .
- Check Sign Change for Inflection Points: We rearrange the equation to isolate the trigonometric function: . Now, we divide both sides by to get . This equation involves both trigonometric and algebraic functions, and it's not straightforward to solve algebraically. We will use a calculator to find the -values that satisfy the equation within the given domain 0 < x < 3.5. We are looking for points where the graph of intersects with the graph of . Using a graphing calculator or a computational tool, we plot both and and look for intersections in the domain 0 < x < 3.5. We find that the -value(s) at the intersection(s) are approximately .
- Check Sign Change for Inflection Points: We rearrange the equation to isolate the trigonometric function: . Now, we divide both sides by to get . This equation involves both trigonometric and algebraic functions, and it's not straightforward to solve algebraically. We will use a calculator to find the -values that satisfy the equation within the given domain 0 < x < 3.5. We are looking for points where the graph of intersects with the graph of . Using a graphing calculator or a computational tool, we plot both and and look for intersections in the domain 0 < x < 3.5. We find that the -value(s) at the intersection(s) are approximately [values to be determined by the calculator]. After finding the -value(s), we need to check if the second derivative actually changes sign around these points to confirm that they are inflection points. We do this by testing values just smaller and just larger than the -value(s) found and observing the sign of at these points.
- Check Sign Change for Inflection Points: We rearrange the equation to isolate the trigonometric function: . Now, we divide both sides by to get . This equation involves both trigonometric and algebraic functions, and it's not straightforward to solve algebraically. We will use a calculator to find the -values that satisfy the equation within the given domain 0 < x < 3.5. We are looking for points where the graph of intersects with the graph of . Using a graphing calculator or a computational tool, we plot both and and look for intersections in the domain 0 < x < 3.5. We find that the -value(s) at the intersection(s) are approximately [values to be determined by the calculator]. After finding the -value(s), we need to check if the second derivative actually changes sign around these points to confirm that they are inflection points. We do this by testing values just smaller and just larger than the -value(s) found and observing the sign of at these points. If the sign of changes from positive to negative or from negative to positive as we pass through the -value(s), then these are indeed inflection points. If the sign does not change, then there is no inflection point at that -value.
