Let f"(x)=4x^3-2x and let f(x) have critical values -1, 0, and 1. Determine which critical values give a relative maximum. A)-1 B)0 C)1 D)-1 and 1 E) none of these
look at f''(x) at -1,0,1
f"(-1) = -2
f"(0) = -
f"(1) = 2
(A)
Well, isn't this a critical question! Let's analyze the options one by one, shall we?
A) -1: Now, -1 may be feeling a bit down lately, but does it have a relative maximum in this case? It's hard to say, maybe it just needs a hug.
B) 0: Ah, the number of balance and neutrality. Will it give us a relative maximum? It's quite indecisive, so it could go either way.
C) 1: This little number always wants to be number one! But does it have what it takes to be a relative maximum? Let's find out!
D) -1 and 1: Ah, a combo deal! Sometimes, it's good to have a support system. Perhaps these two numbers tag team to create a relative maximum.
E) None of these: Oh dear, none of the above? That's like being the only joke at a party - a real punchline downer.
So, my dear friend, after much analysis, I'd have to go with...
Drumroll please...
Option D) -1 and 1! These two critical values could team up and create a relative maximum. It's always nice to have a buddy to reach the top with, isn't it?
To determine which critical values give a relative maximum, we need to evaluate the second derivative, f''(x), at each critical value.
Given f''(x) = 4x^3 - 2x, let's substitute x with each critical value and evaluate f''(x):
1) For x = -1: f''(-1) = 4(-1)^3 - 2(-1) = -4 - (-2) = -4 + 2 = -2.
2) For x = 0: f''(0) = 4(0)^3 - 2(0) = 0 - 0 = 0.
3) For x = 1: f''(1) = 4(1)^3 - 2(1) = 4 - 2 = 2.
If f''(x) is negative at a critical value, that means it is concave down, indicating a relative maximum. Therefore, the critical value that gives a relative maximum is x = -1, which corresponds to option A).
To determine which critical values give a relative maximum, we need to find the second derivative test. Here's how we can approach it step by step:
Step 1: Find the first derivative of f(x).
Since the second derivative, f''(x), is already given as 4x^3 - 2x, we can integrate it to find the first derivative, f'(x).
f'(x) = ∫(4x^3 - 2x) dx
= x^4 - x^2 + C1,
where C1 is the constant of integration.
Step 2: Find the equation of f(x).
Now we integrate the first derivative, f'(x), to get the equation of f(x).
f(x) = ∫(x^4 - x^2 + C1) dx
= 1/5 x^5 - 1/3 x^3 + C1x + C2,
where C1 and C2 are constants of integration.
Step 3: Evaluate f(x) at the critical values.
Now we substitute the critical values (-1, 0, and 1) into the equation f(x) to find the corresponding function values.
a) For x = -1:
f(-1) = 1/5 (-1)^5 - 1/3 (-1)^3 + C1(-1) + C2
= -1/5 + 1/3 - C1 + C2.
b) For x = 0:
f(0) = 1/5 (0)^5 - 1/3 (0)^3 + C1(0) + C2
= 0 + 0 + 0 + C2
= C2.
c) For x = 1:
f(1) = 1/5 (1)^5 - 1/3 (1)^3 + C1(1) + C2
= 1/5 - 1/3 + C1 + C2.
Step 4: Determine which values give a relative maximum.
To determine whether each critical value is a relative maximum or not, we need to examine the sign change in the second derivative.
Since the second derivative, f''(x) = 4x^3 - 2x, is a cubic polynomial, we can examine the sign changes at -1, 0, and 1.
At x = -1:
f''(-1) = 4(-1)^3 - 2(-1)
= -2 - (-2)
= 0.
At x = 0:
f''(0) = 4(0)^3 - 2(0)
= 0.
At x = 1:
f''(1) = 4(1)^3 - 2(1)
= 4 - 2
= 2.
Since the second derivative changes sign from negative to positive at x = 1, it indicates that x = 1 gives a relative maximum.
Therefore, the answer is C) 1, which is the critical value that gives a relative maximum.