Question

I need help with basic limits


Our class was on snow break today and we have homework due tomorrow and we haven't covered the material. the problem is

Find the M(tan) of F(x)=2x^2-x at x=3 give the equation of m(tan)

I know the formula is M(tan) + lim as H approaches 0(f(a+h)-f(a)/h) but I don't know how to plug things in. Thank you!

Answer 1

@DanJS can you please help?

Answer 2

ok

Answer 3

I kinda get the concept, I just have no idea where everything goes

Answer 4

you need to get the derivative from the limit definition for that function?

Answer 5

It's wanting the equation

Answer 6

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

Answer 7

F(x) = 2x^2 - x

\[F'(x) = \lim_{h \rightarrow 0}\frac{ [2*(x+h)^2 - (x+h)] - [2x^2 - x] }{ h }\]

Answer 8

just put in the F(x+h) and the F(x) into that, now expand it all out...

Answer 9

\[\lim_{h \rightarrow 0}\frac{ 2*(x^2+2xh+h^2) -x - h - 2x^2 - x }{ h }\]

Answer 10

\[\lim_{h \rightarrow 0}\frac{ 2x^2+4xh+2h^2 -x - h - 2x^2 - x }{ h }\]

Answer 11

sorry the last x in the numerator should be +x
simplifies more to
\[\lim_{h \rightarrow 0}\frac{ 4xh+2h^2 - h }{ h }\]

Answer 12

factor the h out in the top and it will cancel out with the bottom
\[\lim_{h \rightarrow 0}\frac{ h*(4x+2h - 1) }{ h }\]

Answer 13

\[\lim_{h \rightarrow 0} (4x + 2h - 1) = 4x - 1\]

F ' (x) = 4x - 1

Answer 14

So do I not need the three?

Answer 15

all that was using the limit definition of the derivative to get the derivative function...

F(x) = 2x^2 - x

F ' (x) = 4x - 1

Is M(tan) meaning, the slope of the tangent line at x=3?, never seen the "M(tan)"

Answer 16

Yes, that was just the easiest way to define it. Would I just use slope form?

Answer 17

the derivative function is the slope of a tangent line to F(x) at any given point on the function. The instantaneous rate of change of F at any value of x.

Answer 18

So I plug into the 4(3)-1?

Answer 19

so if x=3

F ' (x) = 4*3 - 1 = 11

The slope of a tangent line to F(x) = 2x^2 - x at he value x=3,
F ' (3) = 11

Answer 20

I've never had to do this before so it's just like a massive overkill. So the equation would be 4(3)-1?

Answer 21

The derivative of F(x) = 2x^2 - x
is
F ' (x) = 4x - 1

that is the slope at any x value on the curve F(x).

When x is 3, the slope of the tangent line there is F ' (3)..

Answer 22

oh, the equation for the tangent line , i see
line with slope F ' (3) = 11, and through point (3 , F(3)) = (3 , 15)

Answer 23

y - y1 = m*(x - x1)

y - 15 = 11*(x - 3)

tangent line

Answer 24

so y-11x-48?

Answer 25

y=11x-48

Answer 26

if you want it in that form ,

y - 15 = 11x - 33
y = 11x - 18

Answer 27

just added the wrong way

Answer 28

Will my basic formula always be the 4xh+2h^2-h/h ?

Answer 29

it depends on the function you have...

The definition is this..
Slope of the tangent line at a point on the curve f(x) --
\[m = F'(x) = \lim_{h \rightarrow 0}\frac{ F(x+h) - F(x) }{ h }\]

Similar to the rise/run idea from algebra, except here we have the limit as the run (change in x value, h) approaches zero. And you end up with the slope at that point tangent to the graph.

Answer 30

This is like the foundation idea for differential calc