Question

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

Answer 1

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!

Answer 2

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.

Answer 3

`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`