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^…

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=

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=

Answer

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})