The following limit represents the derivative of some function f at some number a. limx→5 (2^x−32)/(x−5) State f and a
\[\frac{ f(x)-f(c) }{ x-c }\] thats the formula for the derivative at a certain point.
Can u do this part on your own?
no
may you still help me to get the answer
Well, like I wrote above, you have f(x) - f(c)/x - c. Your problem uses the letter a, but its the same thing f(x) - f(a)/x -a. Now just look at the pjhysical structure of your problem: \[\frac{ f(x)-f(c) }{ x-c }=\frac{ 2x-32 }{ x-5 }\]Kinda see how things match up?
sorry, its 2^x-32/x-5
the x in the 2 is a power
Doesnt change much, really. Still match it up with how the formula looks.
then, a=5, but then f i don't know still
Well, the form of your question fits perfectly with the formula: \[\frac{ f(x)-f(a) }{ x-a }= \frac{ 2^{x} - 32 }{ x-5 }\]So clearly since the formula is x-a on bottom and your problem is x-5, a is 5. So given the formula, where is f(x) positioned? And what is at that same position on your problem?
is 2^x
There ya go. f = 2^x and a = 5
is that the final answer
Yep.
thanks
Of course :)
may you try another one
Which of the following functions satisfy the following conditions? limx→±∞ f(x)=0, limx→3− f(x)=∞, f(2)=0, limx→0 f(x)=−∞, limx→3+ f(x)=−∞. (a) y=(x−2)/x^2(x−3) (b) y=(2−x)/x^2(x−3) (c) y=(2−x)/x^3(x−3)
hello
Do you know how to test any of those? And Im busy, so a bit slow.
yeh, i know but a bit confusing
SO what about the first one? Do you know why lim going to infinity or negative infinity would make the limit 0?
no i dont know that one
Well, if x approaches infinity or negative infinity and the limit ends up being 0, it most of the time means that the denominator was of a higher degree.
okey, thewn will it be c for that one
Actually, that doesnt help us get the answer. C has the highest power int he denominaotr among the 3 choices, but it doesnt matter how high it actually is. If I have a fraction and the fraction has a denominator of a higher degree than the numerator, it becomes 0. SO all 3 of those options would make the limit 0 at infinity. Its just you gotta understand that concept even if it doesnt change much in this case.
okey, understand that one
Okay, then we have limit as x approaches 3 from the left, and we want it to be infinity. This means you want to check a number or two thatis left of x = 3 and plugit into the function and see what happens. If our numbers get increasing high, its infinity, if they get increasingly low, its negative infinity. SO try picking a number or two that is left of 3 and see which functions give you infinity.
so, i think its a
Cant be a. if you check the limit as it goes to 3 from the left, you wont get infinity for letter a.
oh
okey, i must be c
thanks i worked out and find the answer to be b
see you symon
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