use the fact that the derivative of the function g(x)=√x is g(x)=1/(2√x) to find the equation of the tangent…

use the fact that the derivative of the function g(x)=√x is g(x)=1/(2√x) to find the equation of the tangent line to the graph of g(x) at the point x = 81. the equation of the tangent line is y = 1/18x + 9/2. (simplify your answer.)

use the fact that the derivative of the function g(x)=√x is g(x)=1/(2√x) to find the equation of the tangent line to the graph of g(x) at the point x = 81. the equation of the tangent line is y = 1/18x + 9/2. (simplify your answer.)

Answer

Explanation:

Step1: Find the slope of the tangent line

The derivative $g'(x)=\frac{1}{2\sqrt{x}}$. Substitute $x = 81$ into $g'(x)$. So $m=g'(81)=\frac{1}{2\sqrt{81}}=\frac{1}{2\times9}=\frac{1}{18}$.

Step2: Find the y - coordinate of the point

Substitute $x = 81$ into $g(x)=\sqrt{x}$. Then $y = g(81)=\sqrt{81}=9$.

Step3: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(81,9)$ and $m=\frac{1}{18}$. So $y - 9=\frac{1}{18}(x - 81)$.

Step4: Simplify the equation

Expand the right - hand side: $y-9=\frac{1}{18}x-\frac{81}{18}=\frac{1}{18}x-\frac{9}{2}$. Then add 9 to both sides: $y=\frac{1}{18}x-\frac{9}{2}+9=\frac{1}{18}x+\frac{9}{2}$.

Answer:

$y=\frac{1}{18}x+\frac{9}{2}$