# Use calculus to find the slope of the tangent line

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?

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

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.

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)?

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).

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.

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?

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who knows? ... maybe they're both good, and there are two different tangent lines that work.

check out each solution, one at a time.

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?

[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

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?

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