determine all inflection points of f(x)=3(x^3 - 12x^2)ln x + 108x^2 - 7x^3 + 5. as your answer, please enter…

Explanation:

Step1: Find the first - derivative

We use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ and power - rule $(x^n)^\prime=nx^{n - 1}$, $(\ln x)^\prime=\frac{1}{x}$. Let $u = 3(x^{3}-12x^{2})$ and $v=\ln x$. Then $u^\prime=3(3x^{2}-24x)$. The first - derivative of $y = f(x)$ is: [ \begin{align*} f^\prime(x)&=3(3x^{2}-24x)\ln x+3(x^{3}-12x^{2})\frac{1}{x}+216x - 21x^{2}\ &=3(3x^{2}-24x)\ln x+3(x^{2}-12x)+216x - 21x^{2}\ &=3(3x^{2}-24x)\ln x+3x^{2}-36x + 216x - 21x^{2}\ &=3(3x^{2}-24x)\ln x-18x^{2}+180x \end{align*} ]

Step2: Find the second - derivative

Again, use the product rule for the first term. Let $u = 3(3x^{2}-24x)$ and $v=\ln x$. Then $u^\prime=3(6x - 24)$ and $v^\prime=\frac{1}{x}$. [ \begin{align*} f^{\prime\prime}(x)&=3(6x - 24)\ln x+3(3x^{2}-24x)\frac{1}{x}-36x + 180\ &=3(6x - 24)\ln x+9x-72-36x + 180\ &=3(6x - 24)\ln x-27x + 108 \end{align*} ]

Step3: Set the second - derivative equal to zero

[3(6x - 24)\ln x-27x + 108 = 0] [ (6x - 24)\ln x-9x + 36=0] [6(x - 4)\ln x-9(x - 4)=0] [(x - 4)(6\ln x-9)=0]

Step4: Solve for (x)

Case 1: (x - 4=0), then (x = 4) Case 2: (6\ln x-9 = 0), then (\ln x=\frac{3}{2}), and (x = e^{\frac{3}{2}}\approx4.4817)

Answer:

(4 + 4.4817=8.482)