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 of the tangent line 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 the point on the graph of $y = f(x)$ at $x = 4$

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

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_1=m(x - x_1)$, 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 the tangent line is horizontal

A horizontal tangent line has a slope of $0$. Set $f^{\prime}(x)=0$. $\frac{2}{\sqrt{4x - 7}} = 0$. Since the numerator is non - zero ($2\neq0$), there is no solution for $x$ because the fraction $\frac{2}{\sqrt{4x - 7}}$ can never be zero.

Answer:

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