How many critical points does the function

f(x) = [(x-2)^5][(x+3)^4] have?

A. 1
B. 2
C. 3
D. 5
E. 9

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Ok, these are my thoughts on this question:
My impulse was to do derivative of the function. However, if I'm going to set the function equal to zero to find some critical points (because a critical point is where derivative either equals zero or doesn't exist), it's going to take me forever, and this question shouldn't take very long.

So my next choice was to look at the function itself. The degrees are 5 and 4, so if we add those degrees together we get 9. Essentially, I thought the answer would be E because of 9 factors, 9 CPs. However, the factors repeat, 2 and -3!

Thus, my guess is B, which is two. Are my reasonings right?

Thanks much!

lol rice krispies am i right

oh god. didn't even think about factoring. lol thanks

Your reasoning is on the right track! Let's walk through the steps to find the critical points of the given function:

First, let's recall that a critical point is a point where the derivative of the function either equals zero or does not exist.

To find the critical points, we need to find the derivative of the function f(x) and set it equal to zero. However, in this case, taking the derivative of the function directly would be quite long and complicated due to the high degrees of the factors.

But you noticed something important! The factors (x - 2) and (x + 3) are repeated. We can use this to our advantage.

If a factor with multiplicity greater than 1 occurs in a function, it indicates that the graph of the function touches or crosses the x-axis at that particular value multiple times. In this case, the factor (x - 2) has a multiplicity of 5, and the factor (x + 3) has a multiplicity of 4.

Since the factor (x - 2) has a multiplicity of 5, it means that the graph touches or crosses the x-axis at x = 2, and it does so with a higher power of touch or cross. Similarly, the factor (x + 3) indicates that the graph touches or crosses the x-axis at x = -3, and it also does so with a higher power.

Therefore, the function has two critical points, one at x = 2 and another at x = -3. Hence, your reasoning is correct, and the answer is B.

Your first impulse was correct and made sense, but then you slipped into some very fuzzy thinking.

How can you just add the exponents when the bases are not the same??

So from your impulse ....

f'(x)= (x-2)^5(4)(x+3)^3 + (x+3)^4(5)(x-2)^4
=(x-2)^4 (x+3)^3 [ 4(x-2) + 5(x+3)]
= (x-2)^4 (x+3)^3 (9x +7)

x=2, x=-3 , and x = -7/9 will produce critical points.

What was so hard about that?

But you did it wrong