Question

given the function $f(x) = \int_{1}^{x} (t^{4} + 12t^{3} + 36t^{2}) dt$, determine all $x$-values, if any, for which $f$ has a point of inflection.

Explanation:

Step1: Find the first derivative of \( f(x) \)

By the Fundamental Theorem of Calculus, if \( f(x)=\int_{a}^{x}g(t)dt \), then \( f^{\prime}(x) = g(x) \). Here, \( g(t)=t^{4}+12t^{3}+36t^{2} \), so \( f^{\prime}(x)=x^{4}+12x^{3}+36x^{2} \).

Step2: Find the second derivative of \( f(x) \)

Differentiate \( f^{\prime}(x) \) with respect to \( x \). Using the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1} \):
\( f^{\prime\prime}(x)=4x^{3}+36x^{2}+72x \)

Step3: Factor the second derivative

Factor out the common factor \( 4x \) first: \( f^{\prime\prime}(x)=4x(x^{2}+9x + 18) \)
Then factor the quadratic: \( x^{2}+9x + 18=(x + 3)(x+6) \)
So \( f^{\prime\prime}(x)=4x(x + 3)(x + 6) \)

Step4: Find critical points of \( f^{\prime\prime}(x) \)

Set \( f^{\prime\prime}(x)=0 \):
\( 4x(x + 3)(x + 6)=0 \)
This gives \( x = 0 \), \( x=-3 \), and \( x=-6 \)

Step5: Test the sign changes of \( f^{\prime\prime}(x) \)

We need to check the sign of \( f^{\prime\prime}(x) \) in the intervals determined by these critical points: \( (-\infty,-6) \), \( (-6,-3) \), \( (-3,0) \), and \( (0,\infty) \)

• For \( x < -6 \) (e.g., \( x=-7 \)): \( 4(-7)(-7 + 3)(-7+6)=4(-7)(-4)(-1)=-112<0 \)
• For \( -6 < x < -3 \) (e.g., \( x = -4 \)): \( 4(-4)(-4 + 3)(-4+6)=4(-4)(-1)(2)=32>0 \)
• For \( -3 < x < 0 \) (e.g., \( x=-1 \)): \( 4(-1)(-1 + 3)(-1+6)=4(-1)(2)(5)=-40<0 \)
• For \( x>0 \) (e.g., \( x = 1 \)): \( 4(1)(1 + 3)(1+6)=4(1)(4)(7)=112>0 \)

A point of inflection occurs where \( f^{\prime\prime}(x) \) changes sign. We see that \( f^{\prime\prime}(x) \) changes sign at \( x=-6 \), \( x=-3 \), and \( x = 0 \)

Answer:

The \( x \)-values where \( f \) has a point of inflection are \( x=-6 \), \( x=-3 \), and \( x = 0 \)