Finding the Equation of Tangent Line

Find an equation of the tangent line to the graph of f at the indicated point

f(x)=x^{2}+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 x^{2}+2△x+△x^{2}+1-x^{2}-2△x-△x^{2 } ....................................△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 :)

Re: Finding the Equation of Tangent Line

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

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

try again ...

Re: Finding the Equation of Tangent Line

Quote:

Originally Posted by

**Chaim** Find an equation of the tangent line to the graph of f at the indicated point

f(x)=x^{2}+1 at (2, 5)

Here is the setup:

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

Re: Finding the Equation of Tangent Line

Quote:

Originally Posted by

**Plato** Here is the setup:

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

Hmm, so I see you used 2 as the x.

This would simplify it down to (4△x+△x^{2})/△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.

Re: Finding the Equation of Tangent Line

Quote:

Originally Posted by

**skeeter** $\displaystyle f(x) = x^2+1$

$\displaystyle f(x + \Delta x) = (x+\Delta x)^2 + 1$ (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? :o

Thanks

Re: Finding the Equation of Tangent Line

Quote:

Originally Posted by

**Plato** Here is the setup:

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

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

Re: Finding the Equation of Tangent Line

Quote:

Originally Posted by

**Chaim** Hmm, so I see you used 2 as the x.

This would simplify it down to (4△x+△x^{2})/△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.

Quote:

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

Re: Finding the Equation of Tangent Line

Quote:

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? :o

Thanks

Because $\displaystyle f(x)= x^2+ 1$, it goes in **both** f(x+△x) **and** f(x). You added it in one but not the other:

$\displaystyle f(x)= x^2+ 1$ so that $\displaystyle f(x+\Delta x)= (x+\Deltas x)^2+ 1= x^2+ 2x\Delta x+ (\Delta x)^2+ 1$ Then when you subtract the two, the two "1"s will cancel.