Find equations of all tangents to the curve f(x)=9/x that have slope -1

- anonymous

Find equations of all tangents to the curve f(x)=9/x that have slope -1

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- anonymous

At first you have to take the derivative of f. Do you understand why is that?

- anonymous

yes I do, I got -9/x2

- anonymous

Ok, now what is the meaning of the derivative? Knowing this meaning can you guess what the next step would be?

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- anonymous

that is when a function change as the input change, but I'm stuck at the 2nd step
what shall I do next? plug in -1 slope as x?

- anonymous

Yes, in other words it is the rate of change of a function.
But the meaning I was thinking of is the same, but a little diferent.
When the input increases in a really small amount, the value changes too. Since those two changes are very small, the rate of change is approximately the change in the value divided by the increase in the input.
Do you understand what I just said?

- anonymous

not really, can you lead me into solving this? what is the second step I should take?

- anonymous

You need to understand it before gettingg to the next step. But if you want to know, the derivative of a function at x is the slope of its tangent line at this point, that was what I was trying to show you. Then, you should put -9/x2=-1, that is the slope, to find in which points the line passes.

- anonymous

Yes I got to this step, when I was trying to solve for x I got +- 3

- anonymous

but the answer that it should be is y=-x-6 and y = -x+6

- anonymous

so I must have done something wrong right?

- anonymous

Wait, you just found out one point in wich the line passes, it is not the answer yet

- anonymous

What you did so far is correct

- anonymous

oh ic
okay, so the other point is 3 since 9/x = y and x is 3 therefore y = 3 right?

- anonymous

or -3, but yes, thats correct

- anonymous

so now what's next?

- anonymous

Now, you know a point that belongs to the line, and its slope.
The general formula for a line is y=ax+b, and you alrady have a, now put the point you know that belongs to it, and you will find b.

- anonymous

use slope intercept form now

- anonymous

y = mx+b --> -3=-1x+b

- anonymous

sorry, I'm kinda slow at this point, how do I set this = to 9?

- anonymous

You dont need to, you know x and y, from the points (-3, -3) and (3, 3) that this tangent passes through, and you know a. Now:
-3=-1*(-3)+b1
3=-1*3+b2

- anonymous

oh okay

- anonymous

I got it from here

- anonymous

Before, you found out in which points the tangent have this slope, then, you found out the value of the function at this point, and now you use those information to get the formulas for the tangent lines. You seem a little bit confused with what the numbers you are getting mean.

- anonymous

I got it thanks so much for being so patience :)

- anonymous

Your welcome

- anonymous

can I ask you something else?

- anonymous

Of course

- anonymous

same type of problem but it's f(x) = square root of (x+9)

- anonymous

slope 1/2
I did the derivatives and solve for x I got x= -1

- anonymous

I plugged x= -1 back to original equation I got y = +/- 2 sqrt 2
then I use slope intercept to find equations, I got y= 1/4+9/4 sqrt 2

- anonymous

but the answer should be y=1/4+13/4
where did I do wrong?

- anonymous

You got the x wrong.

- anonymous

Or the derivative

- anonymous

original fx is sqrt(x=9)
derivative is 1/2(x+9)^1/2

- anonymous

Where did I do wrong.

- anonymous

its (1/2)(x+9)^(-1/2)

- anonymous

Yes i got that.. But solving for x=-1

- anonymous

Well, thats wrong, did you see the minus, in the exponent?

- anonymous

No i said that mine is 1/{2(x+9)^1/2 same thing

- anonymous

\[\frac{ 1 }{ 2 }=\frac{ 1 }{ 2\sqrt{x+9} }\rightarrow \sqrt{x+9}=1\rightarrow x+9=\pm1\rightarrow x=-10 or -8\]

- anonymous

the slope is 1/4 not 1/2

- anonymous

sorry, didnt ssee you posting.
Then the left side of the equation will be 1/4, and sqrt(x+9)=2 and x+9=+-4, wich gives us: x=-5, and x=-13

- anonymous

Can you go from there?

- anonymous

let me try to do calculation, I did the calculation but got +/- 2 sqrt 2.. give me 1 minute please

- anonymous

Sure

- anonymous

I got it, I think I got messed up when I was trying to square to get rid of sq rt. Thanks
Do you have time to help me on other problems?

- anonymous

I'm almost leaving actually, but I'm here almost every day.

- anonymous

quick question on solving this problem
\[z(4z+7) - x(4x+7) / (z-x)\]

- anonymous

I can combine (z-x) then cancel top and bottom right?

- anonymous

so what is have left is (4z+7) * (4x+7) right?

- anonymous

No, be carefull, think of what is being divided, when you have something that is not being divided you cannot sum them withou a common denominator

- anonymous

oh maybe that's why. thanks

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