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

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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 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
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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