f(x) = 4x3 + 3x2 ? 6x + 5 (a) Find the intervals on which f is increasing. (Ente

Question

f(x) = 4x3 + 3x2 ? 6x + 5 (a) Find the intervals on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection point. (x, y) = Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)

Explanation / Answer

f(x) = 4x^3 + 3x^2 - 6x + 5
f'(x) = 12x^2 + 6x - 6
f''(x) = 24x + 6

(a)

for increasing, f'(x) > 0
12x^2 + 6x - 6 > 0
6(2x^2 + x - 1) > 0
6(2x - 1)(x + 1) > 0
x = (-inf, -1) U (1/2, inf)

for decreasing, f'(x) < 0
12x^2 + 6x - 6 > 0
6(2x^2 + x - 1) > 0
6(2x - 1)(x + 1) > 0
x = (-1, 1/2)

(b)
to find local minimum and maximum value, first we find critical points
put f'(x) = 0
12x^2 + 6x - 6 > 0
x = -1, 1/2
f(-1) = 4x^3 + 3x^2 - 6x + 5 = -4 + 3 + 6 + 5
f(-1) = 10
f(1/2) = 4x^3 + 3x^2 - 6x + 5 = 4/8 + 3/4 - 6/2 + 5
f(1/2) = 1/2 + 3/4 + 2 = 13/4
hence
local minimum value = 13/4
local maximum value = 10

(c)
for inflection points, f''(x) = 0
24x + 6 = 0
x = -1/4

for concave up, f''(x) > 0
24x + 6 > 0
x = (-1/4, inf)

for concave down, f''(x) < 0
x = (-inf, -1/4)

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