Question

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let (f(x)=\frac{x^{4}}{4}+6x^{3}+48x^{2}). a. find the inflection points. enter your answers as ordered pairs. b. identify interval(s) where the graph of (f) is concave up. write none if there is no answer. c. identify interval(s) where the graph of (f) is concave down. write none if there is no answer.

Explanation:

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

Given \(f(x)=x^{4}+6x^{3}+48x^{2}\), using the power rule \((x^n)^\prime = nx^{n - 1}\), we have \(f^\prime(x)=4x^{3}+18x^{2}+96x\).

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

Differentiate \(f^\prime(x)\) with respect to \(x\). \(f^{\prime\prime}(x)=12x^{2}+36x + 96=12(x^{2}+3x + 8)\).

Step3: Find the inflection points

Set \(f^{\prime\prime}(x) = 0\). For the quadratic equation \(x^{2}+3x + 8=0\), the discriminant \(\Delta=b^{2}-4ac\) where \(a = 1\), \(b = 3\), and \(c = 8\). So \(\Delta=3^{2}-4\times1\times8=9 - 32=-23<0\). Since the discriminant is negative, there are no real solutions for \(x\) when \(f^{\prime\prime}(x)=0\), so there are no inflection points.

Step4: Determine concavity

Since \(a = 1>0\) in \(y=x^{2}+3x + 8\) and \(f^{\prime\prime}(x)=12(x^{2}+3x + 8)\), \(f^{\prime\prime}(x)>0\) for all \(x\in(-\infty,\infty)\). The graph of \(f(x)\) is concave up on \((-\infty,\infty)\) and concave down on none.

Answer:

a. NONE
b. \((-\infty,\infty)\)
c. NONE