find all intervals for which the graph of the function y=8x^3-2x^4 is concave downward. find all intervals for which the graph of the function y=8x^3-2x^4 is concave downward. @Mathematics
Use the second derivatie test
f'(x)=24x^2-8x^3 f''(x)=48x-24x^2
alright, that sounds familiar but then what?
find your inflection points by f''(x)=0
f''(x) =0 ?
okay got it, thanks.
so the function is convex for x > x- inflection point
wait what?
yes your inflection points are when f''(x)=0
i got inflection points x=0 and x=2
alright so just like testing for max and mins take a point from each side of your inflection points if f''(c)>0 , it's concave upward f''(c)<0 , concave downwards
so in this case i'd pick c=-1,1,3
yeah that's what i did. I got x<0 and x>2. thanks for the help.
so if you're looking for local minima and maxima, do you use the first or second derivative test?
first
to find the min/max take the crit pt of first derivative and put it in the original function
to get the inflection point take the crit pt of the 2nd derivative and put it in the original eq
okay thanks
lia i forgot this will u plz explain it to me
no problem
explain the question or the solution?
means solution
after finding the first derivative, make f'(x)=0. after finding the solutions to that, use points in between the points that you found to see where the graph of f'(x) will be negative or positive. wherever it's negative, it is concave downward.
thanks
f''(x)=0 is for inflection points or concavity.. Max and mins are for f'(x)
you can also use the 2nd dericative test dont forget. Where you take the crit points of f'(x) and plug them in to f''(x) which is easier way to test for relative extrema
There are all kinds of test.... so yeah pick one and enjoy lol
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