Question

Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing.
f(x)=(x+5)(x+2)(x-7)

Answer 1

Were you able to find the derivative of f(x)? :)

Answer 2

Hey! Do I just multiply all that out and then find the derivative?

Answer 3

Mmm you could, yes.
There is another method though.

Remember the product rule?
It follows the same pattern if we have 3 terms.\[\large (uvw)' \qquad=\qquad u'vw+uv'w+uvw'\]

If that's too confusing we can just multiply it out. :)

Answer 4

I think multiplying out is easier!

Answer 5

I got x^3-39x-70 and the derivative of that would be 3x^2-39

Answer 6

Mmmm yah that sounds right.
Set it equal to zero. Solve for x to find critical points.\[\large 0=3x^3-39\]

Answer 7

Woops that was suppose to be x^2, my bad.

Answer 8

x=6!

Answer 9

or -6 right?

Answer 10

I think we get, \(\large x=\pm \sqrt{13}\)

Answer 11

whoops I did something wrong lol!

Answer 12

adding 39 to each side,

39=3x^2

dividing each side by 3,

13=x^2

:D

Answer 13

Lol I don't know what I was thinking, I think I factored out the 3 or something haha, haha continue!

Answer 14

At our critical points, the function may change from increasing to decreasing, or the other way around.

So we need to pick some test points, and see what's happening on the left and right side of each of our critical points.

The slope of the function is zero at x=-sqrt(13), `because it's a critical point`.

Is the slope negative or positive on the left side of -sqrt(13)?
Let's pick a number slightly smaller (more negative) than it to test.
We'll plug that into our derivative function.

\[\large f'(x)=3x^2-39\]
-sqrt(13) is about -3.6, so let's try x=-4.
\[\large f'(-4)=?\]

Answer 15

9

Answer 16

Ok good. All we care about is the `sign` on our answer it gives us.
Since it's positive, it means our function is `increasing` on this side of our critical point.

Answer 17

Let's try another nice easy number.
x=0 gives us?