Answer: tangent line equation y = (1/5)x + 5
with f '(c) = 1/5
Explanation:
rewrite the function f(x) as f(x) = 2* x^(1/2)
f '(x) = 2* (1/2)* x^(1/2 - 1) = x^(-1/2)
we want f'(c) = f '(25) = (25)^(-1/2) = 1/5
slope of tangent line at x = c is 1/5
y - f(25) = (1/5)* (x - 25)
y - 10 = (1/5)* (x - 25)
y = (1/5)x - 5 + 10
y = (1/5)x + 5