Slope of Tangent Line for g(x)=x^2-4 at (1,-3) | Calculus Homework Solution

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Homework Help Overview

The discussion revolves around finding the slope of the tangent line for the function g(x) = x^2 - 4 at the point (1, -3). Participants are analyzing the limit definition of the derivative to determine the slope at the specified point.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the limit definition of the derivative but are encountering issues with simplification and interpretation of the limit. Questions arise regarding the correctness of their expressions and the handling of indeterminate forms.

Discussion Status

Several participants are providing feedback on the original poster's attempts, suggesting that the limit expression has not been simplified correctly. There is an ongoing exploration of the necessary algebraic manipulations to resolve the limit properly.

Contextual Notes

There is a noted confusion regarding the notation used (g vs. G, f vs. F) and the implications of these notations on the calculations. Participants are also discussing the importance of careful handling of limits, particularly in the context of indeterminate forms.

louie3006
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Homework Statement



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

Homework Equations



lim f(x+Δχ) -F(c)/ (Δχ)

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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louie3006 said:

Homework Statement



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

Homework Equations



lim f(x+Δχ) -F(c)/ (Δχ)

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
[\Delta x ^2 + 2 \Delta x - 3 + 3]/\Delta x

= (\Delta x ^2 + 2 \Delta x)/\Delta x

Factor \Delta x from both terms in the numerator, and cancel with the one in the denominator, then take the limit as \Delta x goes to zero.
 
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 the 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
 
louie3006 said:

Homework Statement



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

Homework Equations



lim f(x+Δχ) -F(c)/ (Δχ)

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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