Given the function $$ f(x)=\int_{1}^{x}\left(t^{4}-16 t^{3}\right) d t $$, determine all $$ x $$-values, if any, for which $$ f $$ has a point of inflection.

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To determine the points of inflection for the function $$ f(x) = \int_{1}^{x} \left(t^{4} - 16t^{3}\right) \, dt $$, we need to analyze the concavity of $$ f $$ by examining its second derivative. ### Step 1: Compute the First Derivative Using the Fundamental Theorem of Calculus, the first derivative of $$ f $$ is: $$ f'(x) = x^{4} - 16x^{3} $$ ### Step 2: Compute the Second Derivative Differentiate $$ f'(x) $$ to find the second derivative: $$ f''(x) = \frac{d}{dx}(x^{4} - 16x^{3}) = 4x^{3} - 48x^{2} = 4x^{2}(x - 12) $$ ### Step 3: Find Critical Points for Inflection Set $$ f''(x) = 0 $$ to find potential inflection points: $$ 4x^{2}(x - 12) = 0 $$ This gives: $$ x = 0 \quad \text{(with multiplicity 2)} \quad \text{and} \quad x = 12 $$ ### Step 4: Analyze the Sign Changes Around Critical Points - **At $$ x = 0 $$:** - For $$ x < 0 $$: $$ f''(x) = 4x^{2}(x - 12) < 0 $$ - For $$ 0 < x < 12 $$: $$ f''(x) = 4x^{2}(x - 12) < 0 $$ There is no sign change around $$ x = 0 $$, so it is **not** a point of inflection. - **At $$ x = 12 $$:** - For $$ x < 12 $$: $$ f''(x) = 4x^{2}(x - 12) < 0 $$ - For $$ x > 12 $$: $$ f''(x) = 4x^{2}(x - 12) > 0 $$ There is a sign change from negative to positive at $$ x = 12 $$, indicating a **point of inflection**. ### Conclusion The function $$ f(x) $$ has a single point of inflection at $$ x = 12 $$. **Answer:** All points of inflection occur at x = 12. The function f has a single inflection point when x = 12. All points of inflection occur at $$ x = 12 $$.

Given the function , determine all -values, if any,
for which has a point of inflection.

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Given the function , determine all -values, if any, for which has a point of inflection.