Question

find all the values of x where the tangent line to the function f(x)=x^3 - 4x^2 - 4x + 9 is horizontal. the solution(s) is/are the value(s) of x that satisfy f(x)=0. that is, solve the equation 3x^2 - 8x - 4. (type an equation.)

Explanation:

Step1: Recall derivative - tangent relation

The slope of the tangent line to a function $y = f(x)$ is given by $f'(x)$. A horizontal tangent line has a slope of 0. So we need to solve $f'(x)=0$. Given $f(x)=x^{3}-4x^{2}-4x + 9$, we first find its derivative using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.
$f'(x)=\frac{d}{dx}(x^{3}-4x^{2}-4x + 9)=3x^{2}-8x - 4$.

Step2: Solve the quadratic equation

We have the quadratic equation $3x^{2}-8x - 4 = 0$. For a quadratic equation $ax^{2}+bx + c=0$ ($a = 3$, $b=-8$, $c = - 4$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-8)^{2}-4\times3\times(-4)=64 + 48=112$.
Then $x=\frac{8\pm\sqrt{112}}{6}=\frac{8\pm4\sqrt{7}}{6}=\frac{4\pm2\sqrt{7}}{3}$.

Answer:

$x=\frac{4 + 2\sqrt{7}}{3},x=\frac{4 - 2\sqrt{7}}{3}$