Question

How can (f(3+h)-f(3))/h equal both 1 and -1? How do I solve for this equation I forgot how to do this!! Please help huge exam is tomorrow!!

Answer 1

im missing something here dude...

(f(3+h)-f(3))/h
\[\frac {f(3+h)-f(3)}{h}\]

is this one of those equations of (as h approaches zero?)

Answer 2

yeah sorry dude

Answer 3

it says as h-->0+ it's 1 and as h-->0- it's -1. How do you find that out and how do you even use algebra to solve this?

Answer 4

do you have the equation for f(x)?

Answer 5

no? isn't that f(x)?

Answer 6

f(x) is any equation (f(x) is just * a function of x is...*
ie f(x) = x^2 + 37

so f(3) = 3^2 + 37
= 9 + 37

Answer 7

i think we may need the equation for f(x) to solve..., hang on I'll ask @satellite73 ...dude u have a sec?

Answer 8

i think it's just a general question, you don't need an equation. I think what they mean to say is for any f(x) it's possible for that to equal 1 and -1

Answer 9

ok...
but assuming : f(x) = x^2 + 37 at x = 3
y = (f(x+h)-f(x))/h
=(( (x+h)^2 + 37)- (x^2 + 37))/h
\[\frac {f(x+h)-f(x)}{h}\]
\[\frac {((x+h)^2 + 37) - (x)^2 + 37}{h}\]
\[\frac {(x^2+2xh+h^2 + 37) - (x)^2 + 37}{h}\] when x = 3
\[\frac {(3^2+2*3*h+h^2 + 37) - (3)^2 + 37}{h}\]
\[\frac {9+6h+h^2 + 37 - 9 - 37}{h}\]
\[\frac {9+6h+h^2 + 37 - 9 - 37}{h}\]
\[\frac {6h+h^2}{h}\]
\[6+h\] so as h --->0
y = 6

i remember now, this is a method for determining a derivative

Answer 10

so if your f(x) is x^2 + 37
f'(x) = 2x
so f'(3) = 2*3 = 6

Answer 11

exactly I don't get it wtf

Answer 12

have you covered derivatives before in class?

Answer 13

all its saying is that at the point x = 3, the slope (gradient) of your equation is either +1 or -1
so your equation is probably something like f(x) = 1/x or 1/sqrt x or something similar