find f(x) and find the equation of the line tangent to the graph of f at the indicated value of x. find the…

Explanation:

Step1: Apply chain - rule

Let $u = 4x - 7$, then $y = u^{\frac{1}{2}}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=4$. So, $f^{\prime}(x)=\frac{1}{2}(4x - 7)^{-\frac{1}{2}}\cdot4=\frac{2}{\sqrt{4x - 7}}$.

Step2: Find the slope at $x = 4$

Substitute $x = 4$ into $f^{\prime}(x)$. $f^{\prime}(4)=\frac{2}{\sqrt{4\times4 - 7}}=\frac{2}{\sqrt{16 - 7}}=\frac{2}{3}$.

Step3: Find $f(4)$

$f(4)=(4\times4 - 7)^{\frac{1}{2}}=\sqrt{16 - 7}=3$.

Step4: Use point - slope form for tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,3)$ and $m=\frac{2}{3}$. So, $y - 3=\frac{2}{3}(x - 4)$, which simplifies to $y=\frac{2}{3}x+\frac{1}{3}$.

Step5: Find where tangent is horizontal

A horizontal tangent line has $f^{\prime}(x)=0$. But $\frac{2}{\sqrt{4x - 7}} = 0$ has no solution since the numerator is non - zero and the denominator $\sqrt{4x - 7}>0$ for the domain of $f(x)$ (where $4x-7>0$).

Answer:

$f^{\prime}(x)=\frac{2}{\sqrt{4x - 7}}$; Tangent line: $y=\frac{2}{3}x+\frac{1}{3}$; No values of $x$ for horizontal tangent.