Post by: ggxxx on April 03, 2011, 04:32:31 pm
I've been preparing for a SAC I have in a couple of days by doing heaps of questions and there is a fair few I'm stuck on... (I'm pretty fair at maths :S). I thought I'd post some here and maybe someone could help me in them. Doesn't have to be all, but even a few Id be grateful :)
Thanks so so much!
1 The average rate of change of the function y = 2 x 3^x over the interval [0, 4] is:
A 40.5
B 162
C 158
D 40
E 4
Given f(x) = x^3 - 2x, the average rate of change of f(x) with respect to x for the interval
[-2, 0] is:
A -2
B 2
C 5
D -3
E 6
(I don't get rates of change!)
The gradient of the chord of the curve y = 2x^2 between the points where x = 1 and
x = 1 + h is given by:
A 2(x + h)^2 - 2x^2
B 4x + 2h
C 4
D 2h
E 4 +2h
7 A golf ball is hit so that its height h(t) metres above the ground t seconds after it is hit is given by h(t) = 3t(8 t). The maximum height reached is:
A 48 metres
B 24 metres
C 8 metres
E 10 metres
E 14 metres
12 The graph of y = | x^2 - 4 | is differentiable for:
A x E R
B x E R+
C x E [-inf, 0]
D R\{-2, 2}
E [-2, 2]
PLEASE HELP! :)
Post by: taiga on April 03, 2011, 04:36:49 pm
Quote
1 The average rate of change of the function y = 2 x 3^x over the interval [0, 4] is:sub in x= 0 and x=4 and record the y values
A 40.5
B 162
C 158
D 40
E 4
x=0, y= 2
x=4, y= 162
Rate of change = change in y/change in x = 160/4 = 40
Hence D
---
Quote
The gradient of the chord of the curve y = 2x^2 between the points where x = 1 and
x = 1 + h is given by:
A 2(x + h)^2 - 2x^2
B 4x + 2h
C 4
D 2h
E 4 +2h
So gradient = rise/run = change in y / change in x
So if we go from x = 1 to x = 1 +h, that means we went up by h, hence the change in x is h.
When x = 1 , y= 2
When x = 1+h, find what y equals, and then write down that value.
Your answer will be = (Final Y Value - 2)/h
Post by: luken93 on April 03, 2011, 04:38:27 pm
Average Rate of Change =
Post by: ggxxx on April 03, 2011, 04:39:18 pm
Thanks heaps! If you don't mind just one more - i have no idea how to do 7 :S the ball one...
Post by: Russ on April 03, 2011, 04:39:54 pm
Recognize that the maximum height will be the turning point of the graph (where the gradient = 0)
Not sure what the weird symbol is but you need to differentiate then set equal to 0 and solve. If you can't differentiate it, post the actual equation and I'll show you
Post by: pi on April 03, 2011, 04:40:51 pm
Quote
12 The graph of y = | x^2 - 4 | is differentiable for:
A x E R
B x E R+
C x E [-inf, 0]
D R\{-2, 2}
E [-2, 2]
D, those two points are cusps
Post by: Water on April 03, 2011, 04:41:03 pm
Post by: ggxxx on April 03, 2011, 04:45:12 pm
The symbol was simply meant to be a "-"... strange :P
Thanks again!
Post by: panicatthelunchbar on April 03, 2011, 06:13:36 pm
So, the easiest method is drawing the graph of the function and then describing it using a hybrid function.
You can then differentiate the hybrid function.
Post by: pi on April 03, 2011, 06:37:23 pm
For question 12, note that you cannot differentiate a modulus function.
So, the easiest method is drawing the graph of the function and then describing it using a hybrid function.
You can then differentiate the hybrid function.
Um, you can. Using the signum functions (although technically, the signum function is also a hybrid function to0). But for that question, you didn't need it.
Post by: panicatthelunchbar on April 04, 2011, 06:26:36 pm
Post by: pi on April 06, 2011, 02:57:44 pm
What's the signum function?
Don't worry, you don't need to know.
http://en.wikipedia.org/wiki/Sign_function