Question

For the function f(x)=2x^3-3x^2-22.5x+19

Answer 1

A. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5\\\\f''(x)=12x-6\\12x-6=0\\12x=6\\x=0.5$ - Relative maximum at x = -1.5. Relative minimum at x = 2.5. Point of inflection at x = 0.5.

B. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5$ - Absolute maximum at x = -1.5. Absolute minimum at x = 2.5. This applies for the interval [-4, 5].

C. $f(x)=2x^3-3x^2-22.5x+19\\f(0)=2(0)^3-3(0)^2-22.5(0)+19\\f(0)=19\\\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=2.5$ - Absolute maximum at x = 0. Absolute minimum at x = 2.5. This applies for the interval [0, 3].

D. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5$ - Test: $f(-2)=2x^3-3x^2-22.5x+19\\f(-2)=2(-2)^3-3(-2)^2-22.5(-2)+19\\f(-2)=2(-8)-3(4)-45+19\\f(-2)=-16-12-26\\f(-2)=-540\\\\f(3)=2x^3-3x^2-22.5x+19\\f(3)=2(3)^3-3(3)^2-22.5(3)+19\\f(3)=2(27)-3(9)-67.5+19\\f(3)=54-27-48.5\\f(3)=-21.5$ - Increasing: $[-\infty,-1.5)(2.5,\infty]$ - Decreasing: $(-1.5,2.5)$

E. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5\\\\f''(x)=12x-6\\12x-6=0\\12x=6\\x=0.5$ - Concave up: $(0.5,\infty]$ - Concave down: $[-\infty,0.5)$