Answer:
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
