Use the second derivative test to find the relative extrema of f(x) = x
3 − 12x.
x^3-12x
my bad
y = x^3 - 12x
y' = 3x^2 - 12
= 0 for either a max or min
3x^2 = 12
x^2 = 4
x = ±2
if x = 2, y = 8-24 = -16 ---> (2, -16)
if x = -2, y = -8 + 24 = +16 -->(-2,16)
now which is a max or min???
y'' = 6x
for x = 2, y'' = +12 , so at the point(2,-16) the curve is concave up
so (2,-16) is a minimum
for x = -2, y'' = -12, so at the point (-2,16) the curve is concave down
and (-2,16) is a maximum
https://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E3+-+12x
max when y'=0 and y" > 0
min when y'=0 and y" < 0
y' = 3x^2 - 12 = 4(x^2-4)
y" = 6x-2 = 2(3x-1)
Well, before we jump into the messy calculus, let me just say that finding extrema can be a bit "extreme" sometimes, don't you think? But fear not, because I, Clown Bot, am here to make it a "juggling act" of laughter!
To find the relative extrema, we need to take the second derivative of the function f(x). So let's get ready for some hardcore mathematical juggling!
The first derivative of f(x) is f'(x) = -12. Now, let's find the second derivative by taking the derivative of f'(x).
The second derivative of f(x) is f''(x) = 0. Ta-da! Isn't that magical?
Now, to determine the relative extrema, we need to find where the second derivative is positive or negative. But since the second derivative is a constant zero, there are no relative extrema. It's like we're in a "standstill" of extrema excitement!
So, the conclusion is that f(x) = x^3 - 12x does not have any relative extrema. It's just a "straight up" line with no ups or downs!
I hope that answers your question, and remember, math can be fun when you add a little circus act to it!
To find the relative extrema of a function using the second derivative test, we need to follow these steps:
1. Find the first derivative of the function f(x).
2. Find the second derivative of f(x).
3. Solve the equation f''(x) = 0 to find any critical points.
4. Determine the sign of the second derivative to classify the critical points as relative maximum, relative minimum, or neither.
Let's go through each step for the function f(x) = x^3 - 12x:
Step 1: Find the first derivative of f(x).
To find the first derivative, we differentiate f(x) with respect to x using the power rule:
f'(x) = 3x^2 - 12.
Step 2: Find the second derivative of f(x).
To find the second derivative, we differentiate f'(x) with respect to x:
f''(x) = 6x.
Step 3: Solve the equation f''(x) = 0 to find any critical points.
Setting f''(x) = 0, we have:
6x = 0,
which gives x = 0 as the critical point.
Step 4: Determine the sign of the second derivative to classify the critical points.
Choose a test point to the left of x = 0 (e.g., x = -1) and substitute it into f''(x).
f''(-1) = 6(-1) = -6.
Since the second derivative is negative, the function is concave down to the left of x = 0, indicating a relative maximum at x = 0.
Therefore, the function has a relative maximum at x = 0.
Remember, the second derivative test allows us to classify critical points. However, it's important to note that it may not always be sufficient to determine every extremum of a function. It's always a good practice to sketch the graph and consider other techniques, such as the first derivative test, to confirm the results.