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There is like two.. mean value for intergrals and derivatives.. you should google that though
If f is continuous on and differentiable on [a,b], then there is c between a and b such that f'(c)=[f(b)-f(a)]/[b-a]
ha umm ok i should just give the question out cause i got confused with that explanation, f(x)= |(x^2-12)(x^2+4)|, how many numbers in the interval -2
find f'(x) plug in c find [f(3)-f(-2)]/[3-(-2)] set that =f'(c) and solve for c
um may be a dumb question but how does one do the derivative of an absolute value?
f(x)=|x|=x when x>0 and -x when x<0 and x=0 when x=0 since the left derivative does not equal to right derivative for f'(0) then the derivative does not exist at 0 f'(x)=1 when x>0 and -1 when x<0
f(x)=|x-2|=x-2 when x>2 and -(x-2) when x<2 and 0 when x=2 f'(x)=1 when x>2 and -1 when x<2 the derivative dne at x=2
f(x)=|x+5|=x+5 when x>-5 and -(x+5) when x<-5 and 0 when x=-5 f'(x)=1 when x>-5 and -1when x<-5 derivative dne at x=-5
f(x)=|2x+5|=2x+5 when x>-5/2 and -(2x+5) when x<-5/2 and 0 when x=-5/2 f'(x)= 2 when x>-5/2 and -2 when x<-5/2 f' dne at x=-5/2
i'm going to get ready for bed but I will come back for a few minutes if you have questions
yeah imma be studying this over so its fine take your time
back for a short time
ummm..ok i did the derivative and its 2x(x^2+4) + (x^2-12)2x but i couldnt get the hang of the absolute value so i left it like that as if it wasnt there
what is the function?
sorry i;m back i had a thing but i will scan what i did
one correction needed one sec i will scan it again
the first one is the wrong problem anyways lol
im tired gn
I was up till 1:30 am still trying to understand this and well i think i got it but in the end we end up having two equations one positive and the other negative right? Also how does one find out if x is less or greater, does it have to be placed on a chart like how you have sqrt of 12 and -12? sorry, slow learner...thank you for your patience