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what is the interval?
is there an interval on which you are to find critical numbers, or there is no interval given >
anyway, I will tell you what to do, before you get to answer me.
the interval is not given
So, you have to: 1) Differentiate the function (take the derivative of the f(x) ). 2) Set f`(x)=0, and the x values that are solutions to this are critical numbers. 3) Critical numbers are also values where f`(x) is undefined - IFF f(x) is defined on that value(s) - but if f(x) is undefined at this value(s) then it is NOT a critical number.
Also, if you were given the interval [a,b] (specifically a closed interval) then a and b would automatically be (also) your critical numbers.
Is the outline/plan clear ?
yes! thank you!
ok, can you find f`(x) for me pliz ?
you can use product rule (or logarithmic differentiation)
working it out now
Product d/dx [ f × g] = f` g + g` h
ok, take your time
i will post the provwe for product rule now.
so i have 8x^7 (x-2)^7 what do i do now?
y = f * g (where f and g are some functions of x) use logarithmic diff. ln(y) = ln(f * g) ln(y) = ln(f) + ln(g) now, f and g are big function, so for any function H, there is a chain rule such that: d/dx [ ln( H(x) ) ] = ( 1 / H(x) ) * H`(x) So here we will use chain for f and g differentiating: (dy/dx) * y=f`/f + g`/g dy/dx = (f`/f + g`/g) * y then look at what y is from the beginning of the problem and substitute. find common denom, and cancel things out
No, that is not correct. don't think so
f(x) = x^8(x − 3)^7 f`(x) = (x^8)` * (x − 3)^7 + x^8 * ( (x-3)^7 )`
f`(x) = 8x^7 * (x-3)^7 + x^8 * 7(x-3)^6
anyway, u got f`(x) then find x values for f`(x)=0 f`(x)=undefined i got to go
ok, thank you!
I got x=0,3 and 24/15
(b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers? At x=? the function has a local max, a local min or neither?