Question

How do you find all critical numbers for the function `f(x)= (x + 2)^3*(x -1)^4`?

Answer 1

Alternatively, we can get the same result by using logarithmic differentiation.

Explanation:

Remember the basic concept of the logarithm: it is the opposite of exponentiation. That is to say, if:

`x^y=z`

Then we can use a logarithm to "cancel out" the exponent, bringing it to the front like so:

`ylnx=lnz`

Now, we don't have to use the natural logarithm (`lnx`) all the time - we could use, say, `log_2(x)` - but mathematicians think the natural log is cool (and it's extremely useful when differentiating, as we'll see in a moment).

Okay, so what's the point? Well, according to the rules of logs, if we take the log of both sides:
`ln(f(x))=ln((x+2)^3(x-1)^4)`

We can split the multiplication into addition:
`ln(f(x))=ln(x+2)^3+ln(x-1)^4`

And remember, exponents come down to the front:
`ln(f(x))=3ln(x+2)+4ln(x-1)`

Now we can take the derivative, making sure to use the chain rule on the left side:
`(f'(x))/f(x)=3/(x+2)+4/(x-1)->` because the derivative of `lnx` is `1/x`

Now just multiply both sides by `f(x)`:
`f'(x)=(3/(x+2)+4/(x-1))f(x)`

And since `f(x)=(x+2)^3(x-1)^4`:
`f'(x)=(3/(x+2)+4/(x-1))((x+2)^3(x-1)^4)`

Finally, set `f'(x)=0` and solve:
`0=(3/(x+2)+4/(x-1))((x+2)^3(x-1)^4)`

We have three solutions:
`0=3/(x+2)+4/(x-1)` and
`0=(x+2)^3` and
`0=(x-1)^4`

The last two clearly mean that `x=-2` and `x=1` are solutions. For the first, we set up the equation:
`-3/(x+2)=4/(x-1)`
`->-3(x-1)=4(x+2)`

Which is a simple equation that yields `x=-5/7`.

So there you have it: same answer, different method. Use which one you feel most comfortable with - that's the beauty of math.