Question

1. find all x - values where the graph of $f(x)=x^{3}+3x^{2}+x + 3$ has a horizontal tangent line.

Explanation:

Step1: Find the derivative

The derivative of $f(x)=x^{3}+3x^{2}+x + 3$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f'(x)=3x^{2}+6x + 1$.

Step2: Set the derivative equal to zero

A horizontal tangent line occurs where the slope of the tangent line (the derivative) is zero. So we set $f'(x)=0$, i.e., $3x^{2}+6x + 1=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 3$, $b = 6$, $c = 1$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substituting the values, we get $x=\frac{-6\pm\sqrt{6^{2}-4\times3\times1}}{2\times3}=\frac{-6\pm\sqrt{36 - 12}}{6}=\frac{-6\pm\sqrt{24}}{6}=\frac{-6\pm2\sqrt{6}}{6}=\frac{-3\pm\sqrt{6}}{3}$.

Answer:

$x=\frac{-3+\sqrt{6}}{3},x=\frac{-3 - \sqrt{6}}{3}$