slope of tangent line


by louie3006
Tags: slope, slope of tangent
louie3006
louie3006 is offline
#1
Feb27-09, 06:30 PM
P: 55
1. The problem statement, all variables and given/known data

find the slope of the tangent line whose g(x)=x^2-4 at point (1,-3)

2. Relevant equations

lim f(x+Δχ) -F(c)/ (Δχ)

3. The attempt at a solution
g(x)= x^2-4
G(1+Δχ)= (1+Δχ)^2-4 ==> Δχ^2+2Δχ-3

lim (Δχ^2+2Δχ-3) - (-3)/(Δχ) = 0
but in the book the answer is 2 so what could I've done wrong?
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Mark44
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#2
Feb27-09, 07:13 PM
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Quote Quote by louie3006 View Post
1. The problem statement, all variables and given/known data

find the slope of the tangent line whose g(x)=x^2-4 at point (1,-3)

2. Relevant equations

lim f(x+Δχ) -F(c)/ (Δχ)

3. The attempt at a solution
g(x)= x^2-4
G(1+Δχ)= (1+Δχ)^2-4 ==> Δχ^2+2Δχ-3

lim (Δχ^2+2Δχ-3) - (-3)/(Δχ) = 0
but in the book the answer is 2 so what could I've done wrong?
Try to keep your letters straight. You have g and G and f and F and x and X and c. There will come a time when this will get you in trouble.

Your last expression (which by the way isn't equal to 0), when simplified a bit, is
[itex][\Delta x ^2 + 2 \Delta x - 3 + 3]/\Delta x[/itex]

= [itex](\Delta x ^2 + 2 \Delta x)/\Delta x[/itex]

Factor [itex]\Delta x [/itex] from both terms in the numerator, and cancel with the one in the denominator, then take the limit as [itex]\Delta x[/itex] goes to zero.
lanedance
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#3
Feb27-09, 07:13 PM
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the limt looks ok until you jump to 0, you still have a deltaX on the denominator, which would tend towrds infinty while the top will tend towards zero. so at teh moment you limit is undetermined until you clean it up a bit more...

so you need to cancel deltaX as much as possible before taking the limit

HallsofIvy
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Feb28-09, 05:45 AM
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slope of tangent line


Quote Quote by louie3006 View Post
1. The problem statement, all variables and given/known data

find the slope of the tangent line whose g(x)=x^2-4 at point (1,-3)

2. Relevant equations

lim f(x+Δχ) -F(c)/ (Δχ)

3. The attempt at a solution
g(x)= x^2-4
G(1+Δχ)= (1+Δχ)^2-4 ==> Δχ^2+2Δχ-3
I assume you mean g(1+ Δx)

lim (Δχ^2+2Δχ-3) - (-3)/(Δχ) = 0
but in the book the answer is 2 so what could I've done wrong?
That limit is NOT 0. If you simply set Δx= 0 you get 0/0 so you have to be more careful.(Δχ^2+2Δχ-3) - (-3)= (Δx)^2+ 2Δx so [(Δχ^2+2Δχ-3) - (-3)]/(Δχ) = (Δx^2+ 2Δx)/Δx= Δx + 2. Take the limit, as Δx goes to 0 of Δx+ 2.


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