How many critical points does the function
f(x) = [(x-2)^5][(x+3)^4] have?
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?
Calculus - Reiny, Monday, May 10, 2010 at 8:29pm
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?
Calculus - Nikki, Monday, May 10, 2010 at 8:35pm
oh god. didn't even think about factoring. lol thanks
Calculus - Bob Pretzel, Monday, February 2, 2015 at 3:17pm
lol rice krispies am i right
Calculus - He ha, Monday, November 16, 2015 at 10:10pm
But you did it wrong