How do you find local maximum value of f using the first and second derivative tests: #f(x) = x^5 - 5x + 5#?

Question

How do you find local maximum value of f using the first and second derivative tests:  #f(x) = x^5 - 5x + 5#?

Christopher Allers · Accepted Answer

#f(-1)=-1+5+5=9" is local maximum"#.

It is known from Calculus that a fun. #f : RRrarrRR# has a local maxima at #x=c", then,":(1): f'(c)=0, &, :(2): f''(c)<0#. #f(x)=x^5-5x+5 rArr f'(x)=5x^4-5, and, f''(x)=20x^3#. #"Now, "f'(x)=0rArr5x^4-5=0rArrx=+-1# Since, #f''(-1)=-20<0,#, we find that, #f(-1)=-1+5+5=9" is local maximum"#.

Christopher Agresta · Answer

undefined To find the local maximum value of $ f(x) = x^5 - 5x + 5 $ using the first and second derivative tests:

1. Find the critical points by setting the first derivative equal to zero and solving for $ x $.
$$ f'(x) = 5x^4 - 5 = 0 $$
$$ x^4 - 1 = 0 $$
$$ (x^2 - 1)(x^2 + 1) = 0 $$
$$ x = \pm 1 $$

2. Evaluate the second derivative at each critical point to determine concavity.
$$ f''(x) = 20x^3 $$
At $ x = -1 $: $ f''(-1) = -20 $, so the concavity is concave down.
At $ x = 1 $: $ f''(1) = 20 $, so the concavity is concave up.

3. Apply the first derivative test:
   - At $ x = -1 $: $ f'(-1) = -5 $, indicating a local maximum.
   - At $ x = 1 $: $ f'(1) = 5 $, indicating a local minimum.

4. Verify using the second derivative test:
   - At $ x = -1 $: $ f''(-1) = -20 < 0 $, confirming a local maximum.
   - At $ x = 1 $: $ f''(1) = 20 > 0 $, confirming a local minimum.

Thus, the local maximum value of $ f(x) $ is at $ x = -1 $, with a maximum value of $ f(-1) = 9 $.