Question

f(x)=x⁴-6x² x ∈ R,Find the maximum and minimum value of the given function.

Answer 1

any function , f(x) attains maximum at x = a only when f'(a) = 0 and f"(a) < 0
similarly, function, f(x) attains minimum at x = b only when f'(b) = 0 and f"(b) > 0

here, f(x) = x⁴ - 6x² , $x\in R$
differentiate f(x) with respect to x,
f'(x) = 4x³ - 12x = 4x(x² - 4)
now, f'(x) = 0
=> 4x(x² - 4) = 0
=> 4x(x - 2)(x + 2) = 0
x = -2, 0 , 2

differentiate once again with respect to x,
f"(x) = 12x² - 12

at x = ± 2 , f"(x) = 12(±2)² - 12 > 0
hence, it attains minimum at x = 2 and -2
so, minimum value of f(x) = x⁴ - 6x²
= (±2)⁴ - 6(±2)² = 16 - 24 = -8

at x = 0, f"(x) = 12 × 0 - 12 < 0
hence, it attains maximum at x = 0
so, maximum value of f(x) = f(0)
= (0)⁴ - 6(0)² = 0

Answer 2

Dear Student:

F(x)=x⁴-6x² , x ∈ R

For,maximum and minimum  

We will find $\frac{df}{dx}$ and then equate to zero.

And then again proceed for second derivative and see it is negative or positive.

If it is positive then f will be minimum

And if negative then f maximum

See the attachment: