In exercise, find the relative extrema of the function.

f (x) = x4 - 2x2 + 5

minima at x = ±1

Step-by-step 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