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Slope of tangent line

  • Thread starter louie3006
  • Start date
1. Homework Statement

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

2. Homework 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?
 
32,580
4,310
1. Homework Statement

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

2. Homework 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

Homework Helper
3,304
2
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

Science Advisor
Homework Helper
41,683
865
1. Homework Statement

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

2. Homework 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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