Question

for which values of x is the slope of the line tangent to the graph of h(x)=\frac{1}{4}x^{4}-\frac{12}{3}x^{3}+8x^{2}-6 horizontal? if there are multiple values, enter them as a comma - separated list. enter dne if no such values exist. x=

Explanation:

Step1: Find the derivative of \(h(x)\)

The derivative of \(h(x)=-\frac{1}{4}x^{4}-\frac{1}{3}x^{3}+8x^{2}-6\) using the power - rule \((x^n)' = nx^{n - 1}\) is \(h'(x)=-x^{3}-x^{2}+16x\).

Step2: Set the derivative equal to 0

We want to find when the slope of the tangent line is 0 (horizontal tangent), so we set \(h'(x)=0\). That is \(-x^{3}-x^{2}+16x = 0\). Factor out \(-x\): \(-x(x^{2}+x - 16)=0\).

Step3: Solve for \(x\)

From \(-x(x^{2}+x - 16)=0\), we have two cases.
Case 1: \(-x=0\), which gives \(x = 0\).
Case 2: Solve \(x^{2}+x - 16=0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b=1\), \(c=-16\). So \(x=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-16)}}{2\times1}=\frac{-1\pm\sqrt{1 + 64}}{2}=\frac{-1\pm\sqrt{65}}{2}\).

Answer:

\(0,\frac{-1+\sqrt{65}}{2},\frac{-1 - \sqrt{65}}{2}\)