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

1 Answer
Aug 13, 2017

Answer:

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.