Consider the equation below. (a) Find the interval on which f is increasing. (En

Question

Consider the equation below.

    (a) Find the interval on which f is increasing.  (Enter your answer in interval notation.)
                            
    
    Find the interval on which f is decreasing.  (Enter your answer in interval notation.)
                            
    
    (b) Find the local minimum and maximum values of f.
         

    (c) Find the inflection points.
    

local minimum          local maximum          Consider the equation below. f(x) = x4 ? 2x2 + 8 Find the interval on which f is increasing. (Enter your answer in interval notation.) Find the interval on which f is decreasing. (Enter your answer in interval notation.) Find the local minimum and maximum values of f.

Explanation / Answer

f(x) = x4 ? 2x2 + 8

f'(x) = 4x^3 - 4x

(a)

for increasing, f'(x) > 0

4x^3 - 4x > 0

4x(x - 1)(x + 1) > 0

i.e. x = (-1, 0) U (1, infinity)

for decreasing, f'(x) < 0

4x^3 - 4x < 0

4x(x - 1)(x + 1) < 0

i.e. x = (-infinity, -1) U (0, 1)

(b)

f''(x) = 12x^2 - 4 = 4(3x^2 - 1)

since critical points are x = -1, 0, 1

at x = 0, f''(x) = -4

at x = -1, f''(x) = 8

at x = 1, f''(x) = 8

hence we have local maximum at x = 0

local minimum at x = -1, 1

(c)

for inflection point, f''(x) = 0

4(3x^2 - 1) = 0

x = -1/sqrt(3), 1/sqrt(3)

at x = -1/sqrt(3), y = 1/9 - 2/3 + 8 = 67/9.............smaller x-value

at x = 1/sqrt(3), y = 1/9 - 2/3 + 8 = 67/9..............larger x-value

(d)

for concave up, f''(x) > 0

4(3x^2 - 1) > 0

x = (-infinity, -1/sqrt(3))

for concave down, f''(x) < 0

4(3x^2 - 1) < 0

x = (1/sqrt(3), infinity)

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