Find the absolute minimum and maximum value of this function. Please help!?

Question

Consider the function f(x) = x^4 - 72 x^2+11, -5 (less than or equal to) x (less than or equal to) 3. 

Find the absolute minimum value of this function. 
Answer: 

Find the absolute maximum value of this function. 
Answer:

Answer 1

The given function is f(x) = x⁴- 72 x²+11

Interval is -5 ≤ x ≤ 3.

Step 1: Find first derivative if f(x)

f(x) = x⁴- 72 x²+11

Differentiate with respect to x.

f'(x) = 4x³- 72*2x

f'(x) = 4x³- 144x

Step 2: Find critical numbers.

To find the critical numbers.

Set f'(x) = 0 then find x values.and Find at which values of x  f'(x) does not exist.

f'(x) = 0 ⇒ 4x³- 144x =0

4x ( x² - 36 ) =0

By using zero product property : If AB = 0 then A = 0 , B = 0.

4x = 0  and ( x² - 36 ) =0

x = 0 and  x = ±6.

x = -6 does not lie in the interval -5 ≤ x ≤ 3, so this is negligible.

x = 6 does not lie in the interval -5 ≤ x ≤ 3, so this is negligible.

x = 0 is lie in the interval -5 ≤ x ≤ 3, so we can consider it as critical number.

Step 3: Test for maxima/minima.

At Left Endpoint:

f(-5) = (-5)⁴- 72 (-5)²+11 = 625 - 360 + 11 = 276 [ Maximum ].

At Critical Number:

f(0) = (0)⁴- 72 (0)²+11 = 11

At Right Endpoint:

f(3) = (3)⁴- 72 (3)²+11 = 81 - 648 + 11 = - 556 [ Minimum ].

Solution :

Absolute Minimum = - 556.

Absolute Maximum = 276.