How many critical points does the function f(x)=(x-2)^5(x+3)^4 have?

3 Answers
Dec 11, 2017

Three

Explanation:

Recall the product and chain rules. These will help you get the derivative without having to make the lengthy expansion of the function.

Chain rule: For a function h(x) = f(g(x)), the derivative is given by h'(x) = f'(g(x)) * g'(x).
Product rule: For a function h(x) =f(x)g(x) the derivative is given by h'(x) = f'(x)g(x) + g'(x)f(x).

Applying these rules:

f'(x) = 5(x - 2)^4(x + 3)^4 + 4(x -2)^5(x + 3)^3

Now we need to let this be 0 and solve for x.

0 = (x - 2)^4(x + 3)^3(5(x + 3) + 4(x- 2))

Thus we have three separate equations:

x -2 = 0
x + 3 = 0
5(x +3) + 4(x- 2) = 0

The solution to the first two equations is immediately visible as being x = 2 and x= -3. The second can be solved as follows.

5(x + 3) + 4(x - 2) = 0

5x + 15 + 4x - 8 = 0

9x = -7

x = -7/9

Therefore, there will be 3 critical points.

Hopefully this helps!

Dec 11, 2017

Three.

Explanation:

f'(x) = 5(x-2)^4(x+3)^4 + (x-2)^5 4(x+3)^3

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

= (x-2)^4(x+3)^3(9x+7)

f'(x) is never undefined and it has 3 zeros.

Dec 11, 2017

color(blue)(x=-3 , x=2, x=-7/9)

Explanation:

If x=c is a critical point of fx) then either of the following are true:

f'(c)=0 or f'(c) does not exist.

First we need to find the derivative of f(x)

d/dx((x-2)^5(x+3)^4)

Using product rule:

f'(a*b)=b*f'(a)+a*f'(b)

If a=(x-2)^5 and b= (x+3)^4

f'(a)=5(x-2)^4*1

f'(b)=4(x+3)^3*1

:.

(x+3)^4*5(x-2)^4*1+(x-2)^5*4(x+3)^3*1

(x+3)^4*5(x-2)^4+(x-2)^5*4(x+3)^3

(x+3)^3(x-2)^4[5(x+3)+4(x-2)]

(x+3)^3(x-2)^4(9x+7)

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

f'(x)=0

(x+3)^3(x-2)^4(9x+7)=0=>color(blue)(x=-3 , x=2, x=-7/9)

All solution are:

-3 ,-3,-3,2,2,2,2,-7/9