Question

question
find all the value(s) of x at which f(x) = 1/1 + 3x^2 - x + 9 has a horizontal tangent line. if there is more than one answer, give all of the x - values separated by commas, e.g. if f(x) has a horizontal tangent line at x = 3 and x = 5 enter 3, 5
provide your answer below:

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=\frac{1}{3}x^{3}+3x^{2}-x + 9\) using the power - rule \((x^n)^\prime=nx^{n - 1}\) is \(f^\prime(x)=x^{2}+6x - 1\).

Step2: Set the derivative equal to zero

A horizontal tangent line occurs when \(f^\prime(x)=0\). So we set \(x^{2}+6x - 1=0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b = 6\), \(c=-1\)), the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) gives \(x=\frac{-6\pm\sqrt{6^{2}-4\times1\times(-1)}}{2\times1}=\frac{-6\pm\sqrt{36 + 4}}{2}=\frac{-6\pm\sqrt{40}}{2}=\frac{-6\pm2\sqrt{10}}{2}=-3\pm\sqrt{10}\).

Answer:

\(-3+\sqrt{10},-3 - \sqrt{10}\)