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

That's "2" not 0. By the way, you should not use "Δχ" as the letter for the increment in "x". It should be "Δx".
  • #1
louie3006
54
0

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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  • #2
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
[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.
 
  • #3
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
 
  • #4
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.
 

1. What is the slope of a tangent line?

The slope of a tangent line is the rate of change of a curve at a specific point. It represents how much the function is changing at that point and is calculated as the slope of the line that touches the curve at that point.

2. How do you find the slope of a tangent line?

To find the slope of a tangent line, you can use the derivative of the function at the point of interest. The derivative represents the slope of the tangent line at any given point on the curve.

3. Can the slope of a tangent line be negative?

Yes, the slope of a tangent line can be negative. It depends on the direction of the curve at the point of interest. If the curve is decreasing at that point, the slope of the tangent line will be negative.

4. What does the slope of a tangent line tell us about a function?

The slope of a tangent line provides information about the rate of change of the function at a specific point. It can tell us if the function is increasing or decreasing at that point and can also be used to find the maximum and minimum points on a curve.

5. Why is the slope of a tangent line important?

The slope of a tangent line is important in calculus and other areas of mathematics because it helps us understand the behavior of a function at a specific point. It is also used in various applications such as determining the speed of an object, finding optimal solutions in optimization problems, and more.

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