Suppose the derivative of a function $f$ is below. On what interval is $f$ increasing?
f'(x) = (x+1)^4(x-5)^3(x-7)^6
Suppose the derivative of a function $f$ is below. On what interval is $f$ increasing?
f'(x) = (x+1)^4(x-5)^3(x-7)^6
$f(x)$ is increasing wherever f'(x) > 0. Thus, it follows that $f(x)$ is increasing wherever:
$(x+1)^4 (x-5)^3 (x-7)^6 > 0$
I hope you can take it from here.
As Tards said:
$f(x)$ is increasing when $f'(x)>0$ which in my opinion should be well known in Calc I.
However, Tards did not answer the question in the form you desired so I will do that.
It is easy to see,
$(x+1)^4(x-5)^3(x-7)^6>0$ when,
$(x+1)^4>0$ which is equivalent to $x>-1$ so the interval is clearly:
$[-1,\infty]$
I know some people may comment and say that I divided the RHS by $((x-5)^3(x-7)^6)$ and if $x=5,7$ I would divide by zero. But if you plug in 5 or 7 it is clearly greater than -1.