Question 1a) 2009 VCAA

[x_anna]

Hi guys, has anyone done this question?

I just did the VCAA 2009 exam and the answer I got was (9,infinity) but their answer was [9,infinity).

I don't understand? The question says 'strictly decreasing'... the stationary point is at x=9 so shouldn't the interval not include 9? :/ I'm pretty sure I've missed something coz VCAA rarely gets anything wrong. Can somebody please explain?

[cranberry]

yeh i got the same as u....I couldn't work out why its including 9

i hate how vcaa solutions don't give a reason for the answer. It's always the answer and then what other students did wrong ..

[ech_93]

I can't explain it very well but ....have a read of this:http://www.atarnotes.com/forum/index.php?topic=125215.0 and http://www.vcaa.vic.edu.au/correspondence/bulletins/2011/April/2011AprilSup2.pdf

[brightsky]

function is strictly decreasing when for every x < y, f(x) > f(y). since it concerns two points, we must look at a specific domain. there is no reason why we need to leave out 9, since 9 < 9 + a, for small positive a, we still have f(9) > f(9 + a), whereas when we take a number lower than 9, say 8, then inequations do not stand.

hope this makes some sort of sense.

[cranberry]

so a can't equal zero?

[brightsky]

hang on, lemme rephrase, letting a=0 doesn't actually achieve anything.

[x_anna]

sighhhh... so confusing. I had a look at VCAA's bulletin (thanks ech ) and I have come to the conclusion that if it says 'strictly increasing/decreasing' I'll include the stationary point value. VCAA and their obscure instructions.

[Asx4Life]

Strictly decreasing means, its decreasing at all subsequent values of x. Meaning As soon as you reach the local maximum. The next value is lower than at the local maximum. Whereas at the local minimum, you don't include the minimum point as the subsequent values are not decreasing anymore.

[x_anna]

Strictly decreasing means, its decreasing at all subsequent values of x. Meaning As soon as you reach the local maximum. The next value is lower than at the local maximum. Whereas at the local minimum, you don't include the minimum point as the subsequent values are not decreasing anymore.

Ooohhh okay, umm I kind of get it now... So if I have a parabola, say y=x^2, it would be strictly decreasing for (-infinity,0) and would be strictly increasing for [0, infinity)?

[Asx4Life]

Strictly decreasing means, its decreasing at all subsequent values of x. Meaning As soon as you reach the local maximum. The next value is lower than at the local maximum. Whereas at the local minimum, you don't include the minimum point as the subsequent values are not decreasing anymore.

Ooohhh okay, umm I kind of get it now... So if I have a parabola, say y=x^2, it would be strictly decreasing for (-infinity,0) and would be strictly increasing for [0, infinity)?

that is correct

[gossamer]

function is strictly decreasing when for every x < y, f(x) > f(y). since it concerns two points, we must look at a specific domain. there is no reason why we need to leave out 9, since 9 < 9 + a, for small positive a, we still have f(9) > f(9 + a), whereas when we take a number lower than 9, say 8, then inequations do not stand.

hope this makes some sort of sense.

I was going to PM you both but it would be better off if I posted to avoid confusion for others.

 Basically, what brightsky has said is correct; the function is strictly decreasing when for every x < y, f(x) > f(y). Since we normally associate x and y with axis, I'm just going to re-write it the way VCAA does in this bulletin: http://www.vcaa.vic.edu.au/correspondence/bulletins/2011/April/2011AprilSup2.pdf (page 3).

A function f is said to be strictly decreasing when a < b implies f(a) > f(b) for all a and b in its domain. A function is said to be strictly decreasing over an interval when a < b implies f(a) > f(b) for all a and b in the interval.

So, if you pick any two points in a given interval, the smaller valued point (which we will call a), will have a greater f(x) value then the bigger valued point (b).

For your example of x^2, it is actually correct to say that the function is strictly decreasing for (-inf,0), because it still fits that definition.

However, what VCAA will probably ask you in an exam (as it did in the '09 exam), is "State the interval for which the graph of  f is strictly decreasing.", and in this case, it requires you to state the maximal interval. Therefore, for y=x^2, you would have to say (-inf,0], because no matter which value lower than 0 that you pick, the corresponding y-value will always be higher than the y-value at 0 itself (which is 0). That's why 0 is able to be included, too.

That was probably a little long-winded, but hopefully it makes sense

[Panicmode]

However, what VCAA will probably ask you in an exam (as it did in the '09 exam), is "State the interval for which the graph of  f is strictly decreasing.", and in this case, it requires you to state the maximal interval. Therefore, for y=x^2, you would have to say (-inf,0], because no matter which value lower than 0 that you pick, the corresponding y-value will always be higher than the y-value at 0 itself (which is 0). That's why 0 is able to be included, too.

Sorry, I have a problem with this. your definition of "A strictly decreasing function"

a < b, f(a) > f(b).

would mean that 0 is not included.

For example taking the function y=x^2

let a = 0, let b = 1

0 < 1 (True)

f(0) >  f(1) false

Therefore, 0 should not be included in the domain. of a function strictly decreasing

[gossamer]

However, what VCAA will probably ask you in an exam (as it did in the '09 exam), is "State the interval for which the graph of  f is strictly decreasing.", and in this case, it requires you to state the maximal interval. Therefore, for y=x^2, you would have to say (-inf,0], because no matter which value lower than 0 that you pick, the corresponding y-value will always be higher than the y-value at 0 itself (which is 0). That's why 0 is able to be included, too.

Sorry, I have a problem with this. your definition of "A strictly decreasing function"

a < b, f(a) > f(b).

would mean that 0 is not included.

For example taking the function y=x^2

let a = 0, let b = 1

0 < 1 (True)

f(0) >  f(1) false

Therefore, 0 should not be included in the domain. of a function strictly decreasing

1 is not in the interval (-infinity,0].

A function is said to be strictly decreasing over an interval when a < b implies f(a) > f(b) for all a and b in the interval.

If b=0, any lower point, a, will work. Therefore 0 is still included.

[Panicmode]

However, what VCAA will probably ask you in an exam (as it did in the '09 exam), is "State the interval for which the graph of  f is strictly decreasing.", and in this case, it requires you to state the maximal interval. Therefore, for y=x^2, you would have to say (-inf,0], because no matter which value lower than 0 that you pick, the corresponding y-value will always be higher than the y-value at 0 itself (which is 0). That's why 0 is able to be included, too.

Sorry, I have a problem with this. your definition of "A strictly decreasing function"

a < b, f(a) > f(b).

would mean that 0 is not included.

For example taking the function y=x^2

let a = 0, let b = 1

0 < 1 (True)

f(0) >  f(1) false

Therefore, 0 should not be included in the domain. of a function strictly decreasing

1 is not in the interval (-infinity,0].

A function is said to be strictly decreasing over an interval when a < b implies f(a) > f(b) for all a and b in the interval.

If b=0, any lower point, a, will work. Therefore 0 is still included.

Ahhhh! Now I get it (=

[x_anna]

Gossamer: wow I actually get it now thankyou!!