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

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At first you have to take the derivative of f. Do you understand why is that?
yes I do, I got -9/x2
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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Other answers:

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?
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?
not really, can you lead me into solving this? what is the second step I should take?
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.
Yes I got to this step, when I was trying to solve for x I got +- 3
but the answer that it should be is y=-x-6 and y = -x+6
so I must have done something wrong right?
Wait, you just found out one point in wich the line passes, it is not the answer yet
What you did so far is correct
oh ic okay, so the other point is 3 since 9/x = y and x is 3 therefore y = 3 right?
or -3, but yes, thats correct
so now what's next?
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.
use slope intercept form now
y = mx+b --> -3=-1x+b
sorry, I'm kinda slow at this point, how do I set this = to 9?
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
oh okay
I got it from here
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.
I got it thanks so much for being so patience :)
Your welcome
can I ask you something else?
Of course
same type of problem but it's f(x) = square root of (x+9)
slope 1/2 I did the derivatives and solve for x I got x= -1
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
but the answer should be y=1/4+13/4 where did I do wrong?
You got the x wrong.
Or the derivative
original fx is sqrt(x=9) derivative is 1/2(x+9)^1/2
Where did I do wrong.
its (1/2)(x+9)^(-1/2)
Yes i got that.. But solving for x=-1
Well, thats wrong, did you see the minus, in the exponent?
No i said that mine is 1/{2(x+9)^1/2 same thing
\[\frac{ 1 }{ 2 }=\frac{ 1 }{ 2\sqrt{x+9} }\rightarrow \sqrt{x+9}=1\rightarrow x+9=\pm1\rightarrow x=-10 or -8\]
the slope is 1/4 not 1/2
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
Can you go from there?
let me try to do calculation, I did the calculation but got +/- 2 sqrt 2.. give me 1 minute please
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?
I'm almost leaving actually, but I'm here almost every day.
quick question on solving this problem \[z(4z+7) - x(4x+7) / (z-x)\]
I can combine (z-x) then cancel top and bottom right?
so what is have left is (4z+7) * (4x+7) right?
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
oh maybe that's why. thanks

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