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…

Explanation:

Step1: Find the slope of the tangent line

The derivative $g'(x)=\frac{1}{2\sqrt{x}}$. Substitute $x = 25$ into $g'(x)$: $g'(25)=\frac{1}{2\sqrt{25}}=\frac{1}{2\times5}=\frac{1}{10}$.

Step2: Find the y - coordinate of the point

Substitute $x = 25$ into $g(x)=\sqrt{x}$. So $g(25)=\sqrt{25}=5$. The point on the curve is $(25,5)$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is the point. Here $m=\frac{1}{10}$, $x_1 = 25$ and $y_1 = 5$. $y-5=\frac{1}{10}(x - 25)$

Step4: Simplify the equation

$y-5=\frac{1}{10}x-\frac{25}{10}$ $y=\frac{1}{10}x-\frac{5}{2}+5$ $y=\frac{1}{10}x+\frac{5}{2}$

Answer:

$y=\frac{1}{10}x+\frac{5}{2}$