How would I find the critical numbers of

f'(x)=(x-3)^3(x+4)(x^2+1)? The answer is x=-4, x=3

I'm assuming that's

f'(x) = (x-3)^3 * (x + 4) * (x^2 + 1)
instead of
f'(x) = (x-3) ^ 3(x + 4)(x^2 + 1)

Critical numbers occur at f'(x) = 0 or does not exist.
Let f'(x) = 0:
0 = (x-3)^3 * (x + 4) * (x^2 + 1)
Will be true if the terms inside the parenthesis are equal to zero (any among them).
x - 3 = 0
x = 3 (critical number)

x + 4 = 0
x = -4 (critical number)

x^2 + 1 = 0
x^2 = -1
x = ±i (imaginary)

thanks!

To find the critical numbers of a function, we need to first find the derivative of the function, and then solve for values of x where the derivative is either zero or undefined.

Let's start by finding the derivative of f(x) = (x-3)^3(x+4)(x^2+1).

To do that, we can use the product rule and the power rule.

1. Using the product rule:
f'(x) = [3(x-3)^2(x+4)(x^2+1)] + [(x-3)^3(x+4)(2x)] + [(x-3)^3(x^2+1)(1)]

2. Simplifying the derivative:
f'(x) = 3(x-3)^2(x+4)(x^2+1) + 2(x-3)^3(x+4)x + (x-3)^3(x^2+1)

Now, to find the critical numbers, we need to set the derivative f'(x) equal to zero and solve for x.

3. Setting f'(x) = 0:
3(x-3)^2(x+4)(x^2+1) + 2(x-3)^3(x+4)x + (x-3)^3(x^2+1) = 0

To solve this equation, we need to factor out common factors.

4. Factoring out (x-3)^2(x+4)(x^2+1) from all terms:
[x-3)^2(x+4)(x^2+1)][3 + 2(x-3)x + 1] = 0

At this point, we have two factors: (x-3)^2(x+4)(x^2+1) and 3 + 2(x-3)x + 1.

5. Setting each factor equal to zero:
First factor: (x-3)^2(x+4)(x^2+1) = 0
Second factor: 3 + 2(x-3)x + 1 = 0

Now we can solve each factor individually:

For the first factor, we have (x-3)^2(x+4)(x^2+1) = 0.

x-3 = 0 -> x = 3
x+4 = 0 -> x = -4
x^2+1 = 0 -> no real solutions, since x^2+1 is always positive

Therefore, x = 3 and x = -4 are critical numbers of f'(x).

For the second factor, we have 3 + 2(x-3)x + 1 = 0.

Simplifying this equation, we get:
2(x-3)x + 4 = 0
2x^2 - 6x + 4 = 0

We can factor out common factors from this quadratic equation:
2(x^2 - 3x + 2) = 0

Factoring the quadratic equation, we get:
2(x-1)(x-2) = 0

Setting each factor equal to zero:
x-1 = 0 -> x = 1
x-2 = 0 -> x = 2

Therefore, x = 1 and x = 2 are additional critical numbers of f'(x).

Combining all the critical numbers, we have:
x = -4, 1, 2, and 3.