Question

In exercise, find the relative extrema of the function.
f(x) = x4 - 2x2 + 5

Answer 1

minima at x = ±1

Explanation:

Given function in the question:

f(x) = x⁴ - 2x² + 5

now,

To find the extrema differentiating the function with respect to 'x' and equate it to zero

we get

f'(x) = 4x³ - (2)(2x) + 0 = 0

or

4x³ - (2)(2x) = 0

or

4x³ - 4x = 0

or

x³ - x = 0

or

x(x² - 1) = 0

or

x = 0 and x = ±1

checking for maxima or minima

again differentiating f'(x) with respect to x, we get

f''(x) = (3)4x² - x

or

f''(x) = 12x² - x

at x = 0

f"(0) = 12(0)² - 0 = 0 [neither maxima nor minima]

at x = -1

f"(-1) = 12(-1)² - (-1) = 12 + 1 = 13 > 0 [relative minima]

at x = 1

f"(1) = 12(1)² - 1 = 12 - 1 = 11 > 0 [relative minima]

hence,

minima at x = ±1