Use calculus to find the slope of the tangent line

In summary, we use derivatives to find the slope of the tangent line at points of tangency A and B on f(x) and g(x). We can use either the original or alternate form of the derivative, and the slopes of the tangent lines must be equal for the two lines to intersect at both points. The relationship between a and b can be expressed as a + b = 1, and solving for b gives two possible values. To find the corresponding values of a, we can plug the values of b into the equation for a + b = 1 and then plug those values of a into f(a) = a^2 to find the coordinates of points A and B.
  • #1
Rawr
15
0
I'm given f(x) = x^2 and g(x) = -x^2 + 2x - 5

And it says

Let A (a, f(a)) be the point of tangency on f(x) and B (b, g(b)) be the point of tangency on g(x). Use calculus to find the slope of the tangent line at each point. What must be true about these two slope expressions?

I know we need to use the derivatives, but do I use the original form of the derivative f(x + deltax) - f(x)/delta x or the alternate form f(x) - f(c)/x-c?

I would think to use the alternate form, because it gives us points of tangency... but when I try to use the alternate form on g(x), I can't seem to get an answer.

For f(x): x^2 - a^2 / (x-a)
(x+a)(x-a) / (x-a)
= x+a

I think I did that right. Now for g(x):
-x^2+2x-5+b^2-2b+5/ (x-b)
-x^2 + 2x + b^2 - 2b/ (x-b)

And I'm basically stuck on what to do from here. Any ideas? Or am I supposed to be using the other derivative equation?
 
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  • #2
f'(c) = lim{x->c} [f(x) - f(c)]/(x - c)

for f(x) = x^2 ...

f'(a) = lim{x->a} [x^2 - a^2]/(x - a)
f'(a) = lim{x->a} [(x + a)(x - a)]/(x - a)
f'(a) = lim{x->a} (x + a) = 2a


for g(x) = -x^2 + 2x - 5 ...

g'(b) = lim{x->b} [(-x^2 + 2x - 5) - (-b^2 + 2b - 5)]/(x - b)
g'(b) = lim{x->b} [b^2 - 2b + 2x - x^2]/(x - b)
g'(b) = lim{x->b} [b^2 - x^2 - 2(b - x)]/(x - b)
g'(b) = lim{x->b} [(b - x)(b + x) - 2(b - x)]/(x - b)
g'(b) = lim{x->b} [(b - x)[(b + x) - 2]]/(x - b)
g'(b) = lim{x->b} [2 - (b + x)] = 2 - 2b
 
  • #3
Ah.. thank you, I get it now. It doesn't matter if you use the alternate form or not, heh.

The next part says use the coordinates of A and B to write an expression for the slope AB in terms of a and b.

So...

f(a) - f(b)/ (a - b)

I think that's what I'm supposed to do. The next part says to substitute things in so there's only one term to solve for. But I don't see what I can substitute.
 
  • #4
something is missing from your original post ... are the two tangent lines to f and g supposed to have some relationship (like being parallel, maybe)?
 
  • #5
skeeter said:
something is missing from your original post ... are the two tangent lines to f and g supposed to have some relationship (like being parallel, maybe)?

Oh, sorry. The graph is supposed to have one tangent line that hits both f(x) and g(x).
 
  • #6
well ... the slopes have to be the same then, don't they?

2a = 2 - 2b
2a + 2b = 2
a + b = 1

now you have the required relationship between a and b.
 
  • #7
Oh! Wow, that answer now, seems so blantant ~_~

So, a = 1-b

Substitute that into the slope equation:

(1-b) ^2 - (-b^2 + 2b -5)
1 - 2b + b^2 + b^2 - 2b + 5
= 2b^2 - 4b + 6

Oookay. Then I need to solve for the equation to get b. The thing is I get two answers so how do I know which one is correct?

To find f(b), I plug it back into 2-2b, right?
 
Last edited:
  • #8
who knows? ... maybe they're both good, and there are two different tangent lines that work.

check out each solution, one at a time.
 
  • #9
Aw man, I just realized I did my math wrong.

For the slope equation, it's supposed to be

2b^2 - 4b + 6 / (1-2b)

How would I solve for b if I'm not given that it equals anything?
 
  • #10
[f(1-b) - f(b)]/[(1-b) - b] = 2 - 2b

[(1-b)^2 - b^2]/(1 - 2b) = 2 - 2b

(1 - 2b)/(1 - 2b) = 2 - 2b

1 = 2 - 2b

2b = 1

b = 1/2

since a + b = 1, a = 1/2
 
  • #11
skeeter said:
[f(1-b) - f(b)]/[(1-b) - b] = 2 - 2b

[(1-b)^2 - b^2]/(1 - 2b) = 2 - 2b

(1 - 2b)/(1 - 2b) = 2 - 2b

1 = 2 - 2b

2b = 1

b = 1/2

since a + b = 1, a = 1/2

Why would it go to b^2? o_o;

It's supposed to be g(b), and g(x) is -x^2 + 2x -5.

lol, probably from my careless typing <.<

*EDIT*

So, I have 2b^2 - 4b + 6 / (1-2b) = 2-2b
2b^2 - 4b + 6 = (2-2b)(1-2b)
2b^2 - 4b + 6 = 2 - 6b + 4b^2
2b^2 - 2b -4 = 0

Factor that out to get

(b - 2) (2b + 2)

b = 2, -1

And then plug those two points in the equation a+b = 1

Solve for a, and then plug those values of a inside f(a) = a^2?
 
Last edited:

1. What is calculus?

Calculus is a branch of mathematics that deals with the study of rates of change and accumulation. It is divided into two main branches: differential calculus and integral calculus.

2. What is the slope of the tangent line?

The slope of the tangent line is the rate of change of a function at a specific point. It represents the instantaneous rate of change, or the steepness of the curve, at that point.

3. How is calculus used to find the slope of the tangent line?

To find the slope of the tangent line using calculus, we use the derivative of the function. The derivative represents the rate of change of the function at any given point, and it can be used to find the slope of the tangent line.

4. What is the process for using calculus to find the slope of the tangent line?

The process involves finding the derivative of the function, which gives us an equation for the slope of the tangent line at any point. We then plug in the x-value of the point we want to find the slope for into this equation to get the slope of the tangent line at that point.

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

The slope of the tangent line is important because it allows us to analyze the behavior of a function at a specific point. It helps us understand the rate of change of the function at that point and can be used to solve many real-world problems in fields like physics, engineering, and economics.

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