find f(x) and find the value(s) of x where the tangent line is horizontal. f(x)=√(x² - 16x + 76) f(x)=□

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{x^{2}-16x + 76}=(x^{2}-16x + 76)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=x^{2}-16x + 76$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. First, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2x - 16$.

Step3: Calculate $f^\prime(x)$

$f^\prime(x)=\frac{1}{2}(x^{2}-16x + 76)^{-\frac{1}{2}}\cdot(2x - 16)=\frac{2x - 16}{2\sqrt{x^{2}-16x + 76}}=\frac{x - 8}{\sqrt{x^{2}-16x + 76}}$.

Step4: Find when the tangent is horizontal

A horizontal tangent occurs when $f^\prime(x)=0$. Since $f^\prime(x)=\frac{x - 8}{\sqrt{x^{2}-16x + 76}}$, we set the numerator equal to 0 (the denominator cannot be 0 for the function to be well - defined). So, $x−8 = 0$, which gives $x = 8$.

Answer:

$f^\prime(x)=\frac{x - 8}{\sqrt{x^{2}-16x + 76}}$; $x = 8$