let F(x) = definite integral b=x^2 a= 2 (24+29sqrt t -4t)dt. classify each critical point of F as a relative maximum, minimum or neither

Answer 1

F(x) = ∫₂^{x^2}(24 + 29 √t - 4t)dt; F'(x) = (24 + 29√x² - 4x²)2x = 0; x = 0 or 24 + 29IxI - 4x² = 0;

4IxI² - 29IxI - 24 = 0; IxI = 8; x = 8 or x = - 8

At x = 0 we have min because derivative changed sign from - to + from left to right.

At x = 8 we have max because derivative change sign from + to -.

At x = - 8 try yourself