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Mathematics 6 Online
OpenStudy (anonymous):

Find f '(x). f(x) =2x^2 ; a=1. Has to be with f(x+h)-f(x)/h formula. Show steps please.

OpenStudy (anonymous):

1. expand 2(x+h)^2

OpenStudy (anonymous):

what do you get?

OpenStudy (anonymous):

2x^2+4xh+2h^2

OpenStudy (anonymous):

subtract 2x^2 from that

OpenStudy (anonymous):

{ f(x+h) -f(x) } done!

OpenStudy (anonymous):

(2h^2 +4xh)/h

OpenStudy (anonymous):

common term?

OpenStudy (anonymous):

h

OpenStudy (anonymous):

2h

OpenStudy (anonymous):

'cancel' it

OpenStudy (anonymous):

2h+4x

OpenStudy (anonymous):

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

OpenStudy (anonymous):

4x

OpenStudy (anonymous):

done:)

OpenStudy (anonymous):

I love you

OpenStudy (anonymous):

I love you too.

OpenStudy (anonymous):

I'm gonna try the next one by myself :D

OpenStudy (anonymous):

Can you answer this? Why is the derivative important?

OpenStudy (anonymous):

It's change. Change is important and sometimes positive.

OpenStudy (anonymous):

Even if it's negative, still good to understand it.

OpenStudy (anonymous):

But couldn't we just find a point close to our original? Seem like the lim and secant line are unnecessary.

OpenStudy (anonymous):

Sure, but 4x is general, it works anywhere... why bother doing calculations for each point when you could just use something that works for every point.

OpenStudy (anonymous):

Right. Well ,I forgot there was a second part to the problem, which is: find the tangent line to the graph of y=f(x) at x=a

OpenStudy (anonymous):

you found the slope of any tangent line at any x (4x)... at x=a =1 the slope of the tangent line is 4*1 a point which the tangent line goes through is (1 , 2(1)^2) so now you know a point on the line and the slope of the line.. can you find the equation for the line?

OpenStudy (anonymous):

omg it is so easy!

OpenStudy (anonymous):

Why didn't I see that?

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

:(

OpenStudy (anonymous):

So let me see if I can figure the line out.

OpenStudy (anonymous):

I'm not getting the right equation. The answer in my book is y=4x-2

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

2 = 4*1 +b b=-2

OpenStudy (anonymous):

I got y=4x

OpenStudy (anonymous):

y=4x -2

OpenStudy (anonymous):

y-4=4(x-1)

OpenStudy (anonymous):

y-2 =4(x-1)

OpenStudy (anonymous):

point on the tangent line is the point on the original function where the tangent line 'hits' it = (1, 2*(1)^2) = (1,2)

OpenStudy (anonymous):

Whoops, I plugged the given number into f '(x) instead of f(x)

OpenStudy (anonymous):

so if a=2 my point would be 2,8, right?

OpenStudy (anonymous):

yep:)

OpenStudy (anonymous):

Okay, just making sure. Thanks for the help!

OpenStudy (anonymous):

sure!

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