Question

Limits and difference quotient:
f(x) = x^2 + x, find f'(3) ...also, find the value of the tangent line for f at x=3

Answer 1

f'(x) (power ule)

Answer 2

Finding the value of the tangent line at x=3 is finding the slope of the function.

Answer 3

Differentiate each term by term.

Answer 4

yes.

Answer 5

So once you find your derivative, plug in x = 3

Answer 6

1) find f'(x)
2) find f'(3) by plugging in 3 for x, into the derivative.
this will be your slope.

then use point slope formula.

Answer 7

yeah.. find f(2), this is the y of the point.

Answer 8

And you'll find the value of the SLOPE of your function at that point. or just the derivative at that point.

Answer 9

Is that what he he looking for, the equation of the tangent line?

Answer 10

perhaps I misinterpreted the question.

Answer 11

he said the "value" which is very ambiguous.

Answer 12

Yeah, I figured.

Answer 13

well, at least he knows what he needs, the slope of the tangent line at x=3, or the equation of it... but we gave the assistance accordingly.

Answer 14

I'm just waiting for the OP to respond.

Answer 15

we're studying limits, and getting into derivatives. I think he's looking for us to use the difference quotient.

Answer 16

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]

Answer 17

so, I'd calculate the limit x->0 using the equation of the slope at the point given and one we create?

Answer 18

f(x)=x^2+x
so
f(x+h)=?

Answer 19

(x+h)^2 + (x + h)

Answer 20

((x+h)^2 + (x + h) - 12) / h

Answer 21

expand (x+h)^2

Answer 22

x^2 + 2xh + h^2

Answer 23

\[f(x) = x^2 + x\]\[f(x+h) = x^2 +2xh + h^2 +x+h\]NOw we need to subtract this by the riginal function.

Answer 24

So
x^2 +2xh + h^2 +x+h - f(x)

= x^2 +2xh +h^2 +x+h - x^2 - x

Answer 25

simplify by combining like terms.

Answer 26

Then you take the simplified version of this function, and divide by h. By taking the limit as h approaches 0, anything that still has an h in it will converge to 0

Answer 27

I came up with an answer of 3, does that sound correct?

Answer 28

Let's check

Answer 29

actually, maybe 2x + 3

Answer 30

2x + 1

Answer 31

x^2 +2xh +h^2 +x+h - x^2 - x

= 2xh +h^2 +h
\[\lim_{x \rightarrow 0} \frac{2xh+h^2+h}{h}\]\[\lim_{x \rightarrow 0} ~(2x+h^2+1)\]\[=2x+1~ \checkmark \]

Answer 32

Good, now what do you have to do?

Answer 33

so, is that the answer to both questions, meaning f prime and the slope of the tangent?

Answer 34

or is f prime what we came up with, and the slope y=2x + 1

Answer 35

oh, this is the equation of your tangent line. \(f'(x) = 2x + 1\).

Now we're evaluating it to find the slope at the point x = 3

Answer 36

oh, and plug in x

Answer 37

yes, what's the slope there?

Answer 38

solve \[f'(3) = 2(3)+1\]

Answer 39

slope is 2?

Answer 40

No, your slope evaluated at x=3 is not 2, unfortunately.

Answer 41

Okay. What we found is the equation of the tangent line itself. The slope of that line is 2.

Your question also asks to find the value of the slope when x = 3 that means we will have a different tangent line equation at that point.

Your question is worded a bit differently than what I am used to, so I was a little confused, sorry.

Answer 42

so, what is the slope when x=3

Answer 43

f'(3) = 2(3)+1 = ?

Answer 44

simplify that expression.

Answer 45

7, so the slope of the tangent line at x=3 is 7

Answer 46

You finally got it :)