Answer to Question #264664 in Real Analysis for Dhruv bartwal

Question

Consider the function f defined on R by

f(x)= 2x^3+ 3x^2-72x- 36

In which of the intervals is the function f increasing, and in which of the intervals is f decreasing? Justify your answer

Answer 1

$Solution: f(x)=2x^3-3x^2-72x-36 \\~~~~~~~~~~~~~~~~~~f'(x)=6x^2-6x-72 \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=6(x^2-x-12) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=6(x^2-4x+3x-12) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=6[x(x-4)+3(x-4)] \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=6(x-4)(x+3) \\~~~~~~~~~~~~~~\therefore f'(x)=0 ~\Rightarrow x=-3,4 \\The ~point~ ~x=-3 ~and~ x=4~ divide ~the ~ real ~ line ~ into ~ three~ disjoint ~intervals. \\<---------|----------|---------> \\- \infty~~~~~~~~~~~~~~~~~~~~~~~~~~~-3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\infty$

In the interval (-∞,-3) and (4,∞), f'(x) is positive while in the interval (-3,4), f'(x) is negative.

Hence, the given function f(x) is increasing in the interval (-∞,-3) and (4,∞) while function f(x) is decreasing in the interval (-3,4).