Lukecrayonz
  • Lukecrayonz
Find the simplified difference quotient of f(x)=sqrt(2x+6)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Lukecrayonz
  • Lukecrayonz
So basically I just need help with sqrt(2(x+h)+6)
Psymon
  • Psymon
\[\frac{ \sqrt{2(x+h)+6}-\sqrt{2x+6} }{ h } \] Multiply top and bottom by the conjugate.
Lukecrayonz
  • Lukecrayonz
[sqrt(2(x+h)+6)-sqrt(2x+6)/h]*h

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Lukecrayonz
  • Lukecrayonz
End answer: 2/sqrt(2x+2h+6)+sqrt(2x+6)
Psymon
  • Psymon
\[\frac{ \sqrt{2(x+h)+6}-\sqrt{2x+6} }{ h }*\frac{ \sqrt{2(x+h)+6}+\sqrt{2x+6} }{ \sqrt{2(x+h)+6} +\sqrt{2x+6}}\] \[\frac{ 2(x+h)+6-2x-6 }{ h(\sqrt{2(x+h)+6}+\sqrt{2x+6}) }\] You get 2x + 2h + 6 - 2x -6, so everything but 2h cancels, but then the h on bottom cancels out with the one on top, so yeah, looks like you got it, lol. Guess I was workign through it but you got it.
Lukecrayonz
  • Lukecrayonz
WAIT OKAY NOW I SERIOUSLY NEED YOUR HELP
Lukecrayonz
  • Lukecrayonz
What in the hell is f '(x)
Lukecrayonz
  • Lukecrayonz
http://gyazo.com/6e93cdf22d480a72829e6023213aa3ba
Psymon
  • Psymon
f'(x) means derivative of f(x)
Lukecrayonz
  • Lukecrayonz
Well then. I have never learned derivatives :O
Psymon
  • Psymon
You have, kinda, but you dont know it. The difference quotient is the "definition" of a derivative. Everytime youve done the difference quotient youve partway found the derivative. To find the derivative, do the differentience quotient and at the very end, make any h's remaining into 0s. What is left after that is the derivative.
Lukecrayonz
  • Lukecrayonz
So do I always turn H into zero, or if it f'(x) lim-->1 I input it as 1?
Lukecrayonz
  • Lukecrayonz
Sorry if thats confusing, its because the first answer is f'(0)
Psymon
  • Psymon
Well, for a derivative its always as h approaches 0. Limit going to a different number is somethign completely different. Also, derivative must be difference quotient in this case. Youll learn how to do derivatives without difference quotient, but this is just the definition. Differene quotient as h goes to 0 = derivative. Anything else is just something else. f'(0) means find the derivative and then set x = 0 after that.
Lukecrayonz
  • Lukecrayonz
Ahhh okay, I thought we set h to zero because of f ' (0)
Lukecrayonz
  • Lukecrayonz
So now the question asks "f'(x)"
Lukecrayonz
  • Lukecrayonz
Is that my final answer of the difference quotient? 5x+2.5h or would it be something else
Lukecrayonz
  • Lukecrayonz
Its 5x :P
Lukecrayonz
  • Lukecrayonz
Well that was easy..
Lukecrayonz
  • Lukecrayonz
The slope of a line tangent to f(x)=x^2 at x=5 should be
Psymon
  • Psymon
Well, thats what you use the derivative for. The derivative gives youa formula for slope. So if you took the derivative of x^2, youd get 2x. Then plug in x = 5, and you see that you get 2(5) = 10, meaning the slope of the tangent line at x = 5 is 10.
Psymon
  • Psymon
Well, derivatives can do a bunch of things, but with graphs it helps give ya slope.
Lukecrayonz
  • Lukecrayonz
Okay wait, what about for f(x)=5x+3
Lukecrayonz
  • Lukecrayonz
5(x+h)+3-(5x+3) 5x+5h+3-5x-3 5h set h to zero 5?
Lukecrayonz
  • Lukecrayonz
So there really isn't any guarantee that theres going to be a variable in the end?
Psymon
  • Psymon
Well, like I just said, the derivative helps you find the slope of a graph. When you want the slope at a certain point, you plug in the x value after you take the derivative. But think of it, this graph is already a line. y = 5x + 3 is slope of 5, y-intercept of 3. So when you take the derivative of course youre going to get 5, the slope on the entire graph is 5.
Lukecrayonz
  • Lukecrayonz
f(x)=2x^2-7x+5
Lukecrayonz
  • Lukecrayonz
2(x+h)(x+h)-7(x+h)+5 2 h^2+4 h x+2 x^2-7x-7h+5-(2x^2-7x+5) -7+2 h+4 x?
Psymon
  • Psymon
Looks good.
Lukecrayonz
  • Lukecrayonz
I added the divide by h in the second step, didn't show work
Lukecrayonz
  • Lukecrayonz
So for my f ' (x), what is my answer?
Psymon
  • Psymon
set h = to 0 now.
Lukecrayonz
  • Lukecrayonz
-7+4x?
Psymon
  • Psymon
Yep, thats the derivative.
Lukecrayonz
  • Lukecrayonz
http://gyazo.com/4971d14bba106e7c898806e9b7f63cdd
Lukecrayonz
  • Lukecrayonz
I'm looking at an example and this just looks so messy.
Psymon
  • Psymon
So an equation for a tangent line is going to be linear. Its going to end up in the old point slope form of y-y1 = m(x-x1) Now we're given a point, so all we need is the slope. Since the derivative is an equation for slope, what we do is take the derivative then plug in the x-coordinate. That will give us the slope we want. Then that slop plus the point go into the point-slope form equation. They need to show you the fast derivative alreayd for these, they take too long otherwise xD
Lukecrayonz
  • Lukecrayonz
You can show me! :D
Psymon
  • Psymon
\[f(x) = x ^{n}\implies f'(x) = nx ^{n-1} \] Basically, the power of comes down as a multiplication and then the power is reduced by 1. So for example, 4x^3. The power of 3 comes down as a multiplication (3)4x^3. And then the power is reduced by 1, so that leaves you with 12x^2. So this problem is x^2 + 5. So the power of x is 2 and comes down to give us 2x^2. Then the power is reduced by 1 giving us 2x. The 5 has x to the power of 0. So when you bring down a 0 power, the 5 goes away and you only have 2x left (constants by themselves are klled in a derivative). So that means the derivative of x^2+5 is just 2x.
Lukecrayonz
  • Lukecrayonz
So wait, take the equation x^2+5, input into differential, and instead of x^2 use 3 for x so I end up with (3+h)^2+5-(3^2+5)/h
Psymon
  • Psymon
No no no, do the derivative first. Once youre done with the derivative and setting h = 0, you get 2x. Plug 3 into THAT one.
Lukecrayonz
  • Lukecrayonz
o.O My book says to do it the way I did. But I'll try your way ^_^ But I got h=-6
Psymon
  • Psymon
h = -6? How did we get h = anything but 0, lol. Its supposed to be limit as h goes to 0.
Lukecrayonz
  • Lukecrayonz
oh sorry I mean m!
Lukecrayonz
  • Lukecrayonz
Okay so the deriv. is 2x
Psymon
  • Psymon
Right. Then plugging in x = 3, since its the x-coordinate of our point, we get m = 6. So now, given an m of 6 and the point (3,14), we can use the point-slope form formula to get y-14 = 6(x-3) y = 6x - 4
Lukecrayonz
  • Lukecrayonz
That method you just taught me is literally the coolest thing i've ever learned in math
Psymon
  • Psymon
Yeah, it lets you do a lot of them really fast. YOull learn it soon. The derivatives do get tricky as the functions get tricky, but these ones we have are all very straight forward.
Lukecrayonz
  • Lukecrayonz
Where do you live?
Psymon
  • Psymon
west coast US.
Lukecrayonz
  • Lukecrayonz
Why are you awake O.O I'm East Coast but still
Psymon
  • Psymon
Dunno. My sleep is always crazy xD Im not up for long, though.
Lukecrayonz
  • Lukecrayonz
Hmm okay, I get the way my book says to do it but I'd like to see your method http://gyazo.com/74c0a9f9eb4d47d9b56aa142fe3f542d
Psymon
  • Psymon
Alrighty. SO -3 - 5x^2. Given the regular formula for derivatives \[x ^{n}\implies nx^{n-1}\] Constants will always become 0 when you take the derivative. The reason is a constant is x^0. And when you bring down the 0 to mulltiply, the whole term is 0. So looking at the second term, I bring the power down as a multiplication to get (2)-5x^2. Then I lower the power by 1 to get me -10x. Because the point is (-4,-83), I plug in this x-coordinate into the derivative meaning my slope at the given point is 40. So using the slope of 40 and the given point, I can put y+83 = 40(x+4) y = 40x+77 So that would be thr tangent line you need.
Lukecrayonz
  • Lukecrayonz
Hmm..
Lukecrayonz
  • Lukecrayonz
Got it:)
Psymon
  • Psymon
Alrighty, awesome. Im off to sleep now, lol. Night
Lukecrayonz
  • Lukecrayonz
NOOOOOOOOOOO:(
Lukecrayonz
  • Lukecrayonz
Gahhh I'll have to find someone else as smart as you
Psymon
  • Psymon
im dead tired, cant stay awake x_x Youll find someone to help. Sorry. Night
Lukecrayonz
  • Lukecrayonz
Thank you so much for your help

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