Finding the Equation of Tangent Line

**Chaim** · October 1st 2012, 02:45 PM

Find an equation of the tangent line to the graph of f at the indicated point
f(x)=x²+1
(2, 5)

This is what I did:
First step - lim (x+△x)²+1-(x+△x)^2)
△x->0 .........................△x
Plugged in 0 to every △x, which turns out to be 1/0, which is undefined

Second step - (x+△x)(x+△x)+1-(x+△x)(x+△x)
....................................△x
Turns into x²+2△x+△x²+1-x²-2△x-△x² ....................................△x
I got lost here, I think I messed up on the 2nd parenthesis, because I thought I should have turned everything negative since there was a minus sign.
But then it cancels everything else except the +1

Can someone help me from here?
Thanks

**skeeter** · October 1st 2012, 02:51 PM

$f(x) = x^2+1$

$f(x + \Delta x) = (x+\Delta x)^2 + 1$ (you forgot the 1)

try again ...

**Plato** · October 1st 2012, 02:59 PM

Originally Posted by Chaim

Find an equation of the tangent line to the graph of f at the indicated point
f(x)=x²+1 at (2, 5)

Here is the setup:
$\lim _{\Delta x \to 0} \frac{{f(2 + \Delta x) - f(2)}}{{\Delta x}} = \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1 - 5}}{{\Delta x}}$

**Chaim** · October 1st 2012, 03:57 PM

Originally Posted by Plato

Here is the setup:
$\lim _{\Delta x \to 0} \frac{{f(2 + \Delta x) - f(2)}}{{\Delta x}} = \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1 - 5}}{{\Delta x}}$

Hmm, so I see you used 2 as the x.
This would simplify it down to (4△x+△x²)/△x
Which then turns into 4x+△x?

I'm not sure if I did it right, the answer at the back of the book is 4x-3.

**Chaim** · October 1st 2012, 04:00 PM

Originally Posted by skeeter

$f(x) = x^2+1$

$f(x + \Delta x) = (x+\Delta x)^2 + 1$ (you forgot the 1)

try again ...

Um, I thought I added the +1 o.o
I used the lim △x->0 (f(x+△x)-f(x))/△x, where this is the definition of a derivative.
Could you explain more on where the +1 goes?

Thanks

**Plato** · October 1st 2012, 04:09 PM

Originally Posted by Plato

Here is the setup:
$\lim _{\Delta x \to 0} \frac{{f(2 + \Delta x) - f(2)}}{{\Delta x}} = \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1 - 5}}{{\Delta x}}$

$\lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1 - 5}}{{\Delta x}}=\lim _{\Delta x \to 0} \frac{4\Delta x + \left( {\Delta x} \right)^2}{{\Delta x}}=\lim _{\Delta x \to 0} {4 + \left( {\Delta x} \right)}$

**HallsofIvy** · October 1st 2012, 05:21 PM

Originally Posted by Chaim

Hmm, so I see you used 2 as the x.
This would simplify it down to (4△x+△x²)/△x
Which then turns into 4x+△x?

No, it doesn't: 4△x/△x= 4, not "4x". Now take the limit as △x goes to 0. That gives the derivative at x= 2 and so the slope of the tangent line at x= 2.

I'm not sure if I did it right, the answer at the back of the book is 4x-3.

**HallsofIvy** · October 1st 2012, 05:25 PM

Originally Posted by Chaim

Um, I thought I added the +1 o.o
I used the lim △x->0 (f(x+△x)-f(x))/△x, where this is the definition of a derivative.
Could you explain more on where the +1 goes?

Thanks

Because $f(x)= x^2+ 1$ , it goes in both f(x+△x) and f(x). You added it in one but not the other:
$f(x)= x^2+ 1$ so that $f(x+\Delta x)= (x+\Deltas x)^2+ 1= x^2+ 2x\Delta x+ (\Delta x)^2+ 1$ Then when you subtract the two, the two "1"s will cancel.

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