first day of calculus. (1 Viewer)

Joined
Sep 13, 2008
Messages
150
Location
St Lucia, Brisbane
Gender
Male
HSC
2010
hey im in year 11 and we just started the wonderful world of calculus yesterday. can you please help me out this this quick question i cant seem to get the right answer which is -13.

f(x) = 3x^2 - 7x - 4 at the point (-1, 6)

Differentiate from first principles to find the gradient of the tangent to the curve.

ive heard you dont use first principle much but it can crop up in your HSC.

thanks for any help if you can show workings.
 

annabackwards

<3 Prophet 9
Joined
Jun 14, 2008
Messages
4,670
Location
Sydney
Gender
Female
HSC
2009
f(x) = 3x^2 - 7x - 4
f(x+h) = 3(x+h)^2 - 7(x+h) - 4

Gradient
= lim [ f(x+h) - f(x) ] / h
(h --> 0)

= lim [3(x+h)^2 - 7(x+h) - 4 - (3x^2 - 7x - 4 ) ] / h
(h --> 0)

= lim [3(x^2+ 2xh + h^2) - 7x -7h - 4 - 3x^2 + 7x + 4 ) ] / h
(h --> 0)

= lim [6xh + 3h^2 -7h] / h
(h --> 0)

= lim h[6x + 3h -7 ] / h
(h --> 0)

= lim [6x + 3h -7 ]
(h --> 0)

= 6x - 7

At (-1, 6), gradient = 6(-1) - 7 = -13
 

annabackwards

<3 Prophet 9
Joined
Jun 14, 2008
Messages
4,670
Location
Sydney
Gender
Female
HSC
2009
No worries.

Remember that you can always check your answer using normal differentiation, which you'll learn soon :)
 

TheStallion

Member
Joined
Mar 24, 2009
Messages
528
Gender
Male
HSC
2010
Yep, learning the first, long method is always a bitch, then when you learn normal differentiation, it all becomes so much easier.
 

jet

Banned
Joined
Jan 4, 2007
Messages
3,148
Gender
Male
HSC
2009
Yeah, it pops up in the HSC, but they always give you the formula.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top