Question

so we found that the slope of the tangent line at x = 36 is $\frac{1}{12}$, and that (36, 6) is a point on the line. substituting these values into the point - slope formula $y - y_0 = m(x - x_0)$, we have $y - y_0 = m(x - x_0)$ $y - 6=\frac{1}{12}(x - 36)$. solving this for y, we can conclude that the equation of the tangent line to $f(x)=sqrt{x}$ at x = 36 is $y=square$.

Explanation:

Step1: Expand the right - hand side

$y - 6=\frac{1}{12}(x - 36)=\frac{1}{12}x-\frac{36}{12}=\frac{1}{12}x - 3$

Step2: Solve for y

Add 6 to both sides of the equation: $y=\frac{1}{12}x - 3+6$

Answer:

$y=\frac{1}{12}x + 3$