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

Aug 13, 2017

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:

$y \ln x = \ln z$

Now, we don't have to use the natural logarithm ($\ln x$) all the time - we could use, say, ${\log}_{2} \left(x\right)$ - 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:
$\ln \left(f \left(x\right)\right) = \ln \left({\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}\right)$

We can split the multiplication into addition:
$\ln \left(f \left(x\right)\right) = \ln {\left(x + 2\right)}^{3} + \ln {\left(x - 1\right)}^{4}$

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

Now we can take the derivative, making sure to use the chain rule on the left side:
$\frac{f ' \left(x\right)}{f} \left(x\right) = \frac{3}{x + 2} + \frac{4}{x - 1} \to$ because the derivative of $\ln x$ is $\frac{1}{x}$

Now just multiply both sides by $f \left(x\right)$:
$f ' \left(x\right) = \left(\frac{3}{x + 2} + \frac{4}{x - 1}\right) f \left(x\right)$

And since $f \left(x\right) = {\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}$:
$f ' \left(x\right) = \left(\frac{3}{x + 2} + \frac{4}{x - 1}\right) \left({\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}\right)$

Finally, set $f ' \left(x\right) = 0$ and solve:
$0 = \left(\frac{3}{x + 2} + \frac{4}{x - 1}\right) \left({\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}\right)$

We have three solutions:
$0 = \frac{3}{x + 2} + \frac{4}{x - 1}$ and
$0 = {\left(x + 2\right)}^{3}$ and
$0 = {\left(x - 1\right)}^{4}$

The last two clearly mean that $x = - 2$ and $x = 1$ are solutions. For the first, we set up the equation:
$- \frac{3}{x + 2} = \frac{4}{x - 1}$
$\to - 3 \left(x - 1\right) = 4 \left(x + 2\right)$

Which is a simple equation that yields $x = - \frac{5}{7}$.

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