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Suppose the derivative of a function $f$ is below. On what interval is $f$ increasing?

$$ f'(x) = (x+1)^4(x-5)^3(x-7)^6 $$

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Where does the second derivative come in? – Arturo Magidin Nov 28 '11 at 2:10
Why use the Second Derivative Test ( The hint/answer provided below should answer your questions. – JavaMan Nov 28 '11 at 2:14

$f(x)$ 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.

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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:


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.

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