Find all relative extrema of the function. (If an answer does not exist, enter DNE.) f(x) = x^2 + 2x + 14
"extrema" as in minima and maxima?
If so, take the first derivative and set it equal to zero. Find all the solutions (values of x where this is true). With the original function, determine whether these points are minima or maxima by taking points just before and just after.
so if I take the first derivative it would be f'=2x+2 and then setting it to zero would look like f'=2x+2=0, then if x=-1 it would then equal 0. So would I then take the limit as it approaches x at -1 for f(-1)=x(-1)^2+2(-1)+14?
im not sure if this is what the question is askin for, is it this? |dw:1406496364992:dw|
I'm not sure what extrema is.. let me see what else i can find out...
This is what it looks like
it looks like it is asking for 4 points
would I then put -1, 0 and then 0, -1?
no, thats not right. I'm not sure, look at the graph http://www.wolframalpha.com/input/?i=x%5E2%2B2x%2B14
theres only 1 minimum, the maxima at both ends is infinity
soo looking at the link it says global minimum is -1? I'm not sure how to input it for (x,y). That means x=-1 and y=0 right? and then the maxima would be 0,inf?
it would be ( -1, f(-1) ) f(-1) is what you found above
then maxima \(\pm \infty\), not sure how you're supposed to format it
no, i dont open word files from the internet.
sure, or you can post it as a pdf, which would be quicker
relative maximum DNE relative min: (-1, 13)
so there is no maximum? and for the bottom two i just had to solve for it? that isn't so bad. :)
it's not eh! :) the main point of this is to take the derivative and find where f'(x)=0, those are the "turning points" of the graph, where the slope is horizontal. Then determine whether those points are maxima or minima.
oh ok! Thank you for explaining it to me. I hope it sticks long enough to take my next test! :)
haha write it down! :P but no problem!
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