Answer:
f(5) = 2
f'(5) = [tex]\frac{1}{5}[/tex]
Step-by-step explanation:
Tangent line to a function y = f(x) on a point (5, 2) passes through two points (5, 2) and (0, 1)
Let the equation of the line is,
y - y' = m(x - x')
Slope of a line passing through [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
= [tex]\frac{2-1}{5-0}[/tex]
= [tex]\frac{1}{5}[/tex]
Therefore, equation of the line passing through (0, 1) and slope = [tex]\frac{1}{5}[/tex] will be,
y - 1 = [tex]\frac{1}{5}(x-0)[/tex]
y = [tex]\frac{x}{5}+1[/tex]
Function representing equation will be,
f(x) = [tex]\frac{x}{5}+1[/tex]
At x = 5,
f(5) = [tex]\frac{5}{5}+1[/tex]
= 1 + 1
= 2
f(5) = 2
f'(x) = [tex]\frac{d}{dx}(\frac{x}{5}+1)[/tex]
= [tex]\frac{1}{5}[/tex]
Therefore, f'(5) = [tex]\frac{1}{5}[/tex] will be the answer.
