Relative Extrema: Graphing f(x) = (x + 1)^4

[airforceone]

Suppose I am to draw a quick graph of: (x + 1)^4

So I see that zeros are: { -1 (mult 4) }

and: Relative extremas = 3

When I graphed, I drew this:

But when I entered the equation in the calculator, it shows this:

My graph shows the 3 relative extremas, but the calculator doesn't?

Thanks!
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Edited by stapel -- Reason for edit: Moving question to appropriate forum.

 

[o_O]

Re: Relative Extrema Graphing Help

Mmm ... not sure I follow what you did there. How did you figure out there were 3 extrema?

To find the relative/absolute extrema, you set f'(x) = 0:

\(\displaystyle f(x) = (x+1)^{4}\)

\(\displaystyle f'(x) = 4(x+1)^{3} = 0\)

If you solve for x, you can see that there is only one candidate for a relative/absolute extremum.

Edit: Sorry. I'm assuming you know some calculus?

 

[skeeter]

the graph of \(\displaystyle y=(x+1)^4\) is just the graph of \(\displaystyle y=x^4\) shifted left 1 unit.