find f(x) and find the value(s) of x where the tangent line is horizontal.\nf(x) = \\sqrt{x^{2}-14x +…

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{x^{2}-14x + 51}=(x^{2}-14x + 51)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=x^{2}-14x + 51$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{d}{du}(u^{\frac{1}{2}})=\frac{1}{2}u^{-\frac{1}{2}}$. Second, find $\frac{du}{dx}$: $\frac{d}{dx}(x^{2}-14x + 51)=2x-14$. Then $f^\prime(x)=\frac{1}{2}(x^{2}-14x + 51)^{-\frac{1}{2}}\cdot(2x - 14)$.

Step3: Simplify the derivative

$f^\prime(x)=\frac{2x-14}{2\sqrt{x^{2}-14x + 51}}=\frac{x - 7}{\sqrt{x^{2}-14x + 51}}$.

Step4: Find when the tangent line is horizontal

A horizontal tangent line occurs when $f^\prime(x)=0$. Since $f^\prime(x)=\frac{x - 7}{\sqrt{x^{2}-14x + 51}}$, and the denominator $\sqrt{x^{2}-14x + 51}\neq0$ for the function to be well - defined, we set the numerator equal to 0. So $x-7 = 0$, which gives $x = 7$.

Answer:

$f^\prime(x)=\frac{x - 7}{\sqrt{x^{2}-14x + 51}}$; $x = 7$