Let f'' (x)= 4x^3 - 2x and let f(x) has critical numbers -1, 0, and 1, how do you use the second derivative test to determine which critical numbers, if any give a relative maximum?

Question

Elijah Abram · Accepted Answer

For each critical number #c#, find #f''(c)#.
If #f''(c) >0#, then #f(c)# is a relative minimum,
if #f''(c) < 0#, then #f(c)# is a relative maximum,
#f''(c) = 0#, then the test fails (We would use the first derivative test if we knew #f'(x)#.) For example:  #f''(-1) = 4(-1)^3-2(-1) =-4 +2 < 0# so #f(-1)# is a relative maximum. ( Note: You can say more about the critical number #0# in this problem after you know how to "work backwards" from #f''(x)# to #f'(x)#.)

Elijah Abram · Answer

undefined To use the Second Derivative Test to determine which critical numbers give a relative maximum, follow these steps: 1. Find the critical numbers of $f(x)$ by setting its first derivative equal to zero and solving for $x$. In this case, you're given that the critical numbers are -1, 0, and 1. 2. Evaluate the second derivative $f''(x)$ at each critical number to determine its concavity. 3. If $f''(x) > 0$ at a critical number $x$, then $f(x)$ is concave up at that point, indicating a relative minimum. 4. If $f''(x) < 0$ at a critical number $x$, then $f(x)$ is concave down at that point, indicating a relative maximum. 5. If $f''(x) = 0$, the test is inconclusive. So, for each critical number (-1, 0, and 1): - Evaluate $f''(-1)$, $f''(0)$, and $f''(1)$. - If $f''(x) < 0$ at any critical number, it corresponds to a relative maximum. - If $f''(x) > 0$ or $f''(x) = 0$, then it does not correspond to a relative maximum. Apply this process to determine which critical numbers, if any, give a relative maximum.

Emily Ammerman · Answer

undefined To use the second derivative test to determine which critical numbers, if any, give a relative maximum, follow these steps:

1. Find the second derivative of the function $f(x)$.
2. Evaluate the second derivative at each critical number.
3. If the second derivative is positive at a critical number, then the function has a relative minimum at that point.
4. If the second derivative is negative at a critical number, then the function has a relative maximum at that point.
5. If the second derivative is zero or undefined at a critical number, the test is inconclusive, and other methods may be needed to determine the nature of the critical point.

Given $f''(x) = 4x^3 - 2x$, we can evaluate this second derivative at each critical number: -1, 0, and 1.

1. For $x = -1$:
$$f''(-1) = 4(-1)^3 - 2(-1) = 4 + 2 = 6$$
Since $f''(-1)$ is positive, the function $f(x)$ has a relative minimum at $x = -1$.

2. For $x = 0$:
$$f''(0) = 4(0)^3 - 2(0) = 0$$
The second derivative at $x = 0$ is zero. The second derivative test is inconclusive at this point.

3. For $x = 1$:
$$f''(1) = 4(1)^3 - 2(1) = 4 - 2 = 2$$
Since $f''(1)$ is positive, the function $f(x)$ has a relative minimum at $x = 1$.

Therefore, according to the second derivative test, the critical numbers that give a relative maximum are -1.