# Consider the function f(x) = −2x^3 + 3x^2 + 36x?

### 3 Answers

- MyRankLv 64 weeks ago
f(x) = -2x³ + 3x² + 36x

-2x³ + 3x² + 36xf’(x) = -2 x 3x² + 3 x 2x + 36= -6x² + 6x + 36Put x = 1, 2, 3, …………f’(1) = -6(1)² + 6(1) + 36= -6 + 6 + 36 > 0Increasing function f(x)f’(x) = 0-6x² + 6x + 36x² - x - 6 = 0x² - 3x + 2x - 6 = 0x(x - 3) + 2(x - 3) = 0(x - 3) (x + 2) = 0x = 3, -2intervals x ϵ [3, -2]f’(x) = 0 function is local minimumf’(x) > 0inflection points [3, -2].

Source(s): http://myrank.co.in/ - TomVLv 74 weeks ago
f(x) = -2x³ + 3x² + 36x

a) Find the intervals in which f is increasing or decreasing.

f(x) increases where f'(x) > 0

f'(x) = -6x² + 6x + 36 > 0

6x² - 6x - 36 < 0

x² - x - 6 < 0

(x-3)(x+2) < 0

If x - 3 > 0, x+2 < 0

{x > 3}∩{x < -2} = {null}

No solution

if x - 3 < 0, x+2 > 0

{x < 3}∩{x > -2} = {-2 < x < 3}

f(x) decreases where f'(x) < 0

f'(x) = -6x² + 6x + 36 < 0

x² - x - 3 > 0

(x-3)(x+2) > 0

if (x-3) > 0, (x+2) > 0

{x > 3}∩{x > -2} = {x > 3}

If (x-3) < 0, (x+2) < 0

{x < 3}∩{x < -2} = {x < -2}

Ans:

f(x) increases on -2 < x < 3

f(x) decreases on {x > 3}U{x < -2} = (-∞, -2)U(3, ∞)

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b) Find the intervals of concavity

f"(x) = -12x + 6

f(x) is concave upward when f"(x) > 0

-12x + 6 > 0

x < 1/2

f(x) is concave downward when f"(x) < 0

-12x + 6 < 0

x > 1/2

Ans:

f(x) is concave up on x < 1/2

f(x) is concave downward on x > 1/2

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c) Local extrema occur where f'(x) = 0

f'(x) = -6x² + 6x + 36 = 0

x² - x - 6 = 0

(x-3)(x+2) = 0

x = -2, 3

f(x) is maximum at f'(x) = 0 and f"(x) < 0

f(x) is minimum at f'(x) = 0 and f"(x) > 0

f"(-2) = 30 > 0

f"(3) = -30 < 0

Ans:

(-2, f(-2)) is a local minimum

(3, f(3)) is a local maximum

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d) Find the inflection points.

Inflection points occur where f(x) is defined and f"(x) = 0 at the boundary of two intervals of opposite concavity.

Ans: (1/2, f(1/2)) is an inflection point

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- Anonymous4 weeks ago
I want to find the intervals of cockackivity and some poles with it, alright.

thats fine