Question

(5) 7. give the equation of the tangent to y = √x + 2/√x at x = 4.

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{x}+\frac{2}{\sqrt{x}}$ as $y=x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}$.

Step2: Find the derivative

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $y'=\frac{1}{2}x^{-\frac{1}{2}}-x^{-\frac{3}{2}}$.

Step3: Find the slope of the tangent at $x = 4$

Substitute $x = 4$ into $y'$. $y'(4)=\frac{1}{2}(4)^{-\frac{1}{2}}-(4)^{-\frac{3}{2}}=\frac{1}{2}\times\frac{1}{2}-\frac{1}{8}=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$.

Step4: Find the y - coordinate at $x = 4$

Substitute $x = 4$ into $y$. $y(4)=\sqrt{4}+\frac{2}{\sqrt{4}}=2 + 1=3$.

Step5: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,3)$ and $m=\frac{1}{8}$. So $y - 3=\frac{1}{8}(x - 4)$.

Step6: Simplify the equation

$y-3=\frac{1}{8}x-\frac{1}{2}$, then $y=\frac{1}{8}x+\frac{5}{2}$.

Answer:

$y=\frac{1}{8}x+\frac{5}{2}$