Author Topic: 1,000,000 Question Thread :D (Read 45235 times) Tweet Share

kenhung123 · « **Reply #75 on:** December 06, 2009, 03:22:05 pm »

Is the image correct?

TrueTears · « **Reply #76 on:** December 06, 2009, 03:24:37 pm »

Quote from: kenhung123 on December 06, 2009, 01:03:47 pm

Find: y<x^2+4x+4
Is there an actual algebraic answer or just a shaded graph?

Okay if it said solve for x...

$(x+2)^2 > y$

Case 1:

$x+2 > 0$

$x > \pm \sqrt{y} - 2$

Case 2:

$x+2<0$

$-(x+2) > \pm \sqrt{y}$

$x+2 < \pm \sqrt{y}$

$x < \pm \sqrt{y} - 2$

kenhung123 · « **Reply #77 on:** December 06, 2009, 03:33:41 pm »

Thanks

kenhung123 · « **Reply #78 on:** December 06, 2009, 03:34:47 pm »

Is the root of a quadratic (a perfect square) a modulus function?

TrueTears · « **Reply #79 on:** December 06, 2009, 03:44:06 pm »

Quote from: kenhung123 on December 06, 2009, 03:34:47 pm

Is the root of a quadratic (a perfect square) a modulus function?

$y = (x+2)^2$

$x = -2$

$x = -2$ is not a modulus function.

kenhung123 · « **Reply #80 on:** December 06, 2009, 03:55:52 pm »

Let a be a positive number, let $f:[2,\propto ]\rightarrow R, f(x)=a-x$ and let $g:(-\propto ,1]\rightarrow R, g(x)=x^2+a$ Find all values of a for which f o g and g o f both exist.

TrueTears · « **Reply #81 on:** December 06, 2009, 03:58:52 pm »

Quote from: kenhung123 on December 06, 2009, 03:55:52 pm

Let a be a positive number, let $f:[2,\propto ]\rightarrow R, f(x)=a-x$ and let $g:(-\propto ,1]\rightarrow R, g(x)=x^2+a$ Find all values of a for which f o g and g o f both exist.

$ran f \subseteq (- \infty, 1]$ , $ran g \subseteq [2, \infty)$

So 1. we need to find $a$ so that any value of $f:[2 ,\infty) \rightarrow R, f(x)=a-x$ will only be a subset of $(-\infty, 1]$

The function $f(x)$ is a straight line and always decreasing, so the maximum value will be at its left endpoint, at $f(2) = a-2$ . Hence, we can say $ranf = (-\infty, a-2]$ . For it to have the correct range, we require $a-2 \leq 1 \Rightarrow a \leq 3$

and 2. we need to find $a$ so that any value of $g:(-\infty, 1] \rightarrow R, g(x) = x^2+a$ will only be a subset of $[2, \infty)$

For $g(x) = x^2+a$ , $a$ determines the vertical translation of the standard parabola with vertex at (0,0). We can say $rang = [a, \infty)$ . So to have the right range, $a \geq 2$ .

Hence the answer is $2 \leq a \leq 3$ .

btw code for infinity is \infty

TrueTears · « **Reply #82 on:** December 06, 2009, 03:59:29 pm »

lol I posted this question earlier in this thread as an exercise

Quote from: TrueTears on December 05, 2009, 07:09:26 pm

Here's a good composite functions exercise and will really test your understanding. From memory I think it is from Essentials.

Let $a$ be a positive number. Let $f : [2, \infty) \to \mathbb{R}, f(x)=a-x$ and let $g:(-\infty,1] \to \mathbb{R}, g(x)= x^2 +a$ . Fine all values of ' $a$ ' for which $f(g(x))$ and $g(f(x))$ both exist

kenhung123 · « **Reply #83 on:** December 06, 2009, 04:12:11 pm »

Thanks great explanations. May I ask how do you just know how to solve these problems? I can't understand how to solve problems the way they are supposed to be. Nor is there any conventions to follow.

TrueTears · « **Reply #84 on:** December 06, 2009, 04:12:49 pm »

Quote from: kenhung123 on December 06, 2009, 04:12:11 pm

Thanks great explanations. May I ask how do you just know how to solve these problems? I can't understand how to solve problems the way they are supposed to be. Nor is there any conventions to follow.

Practise, try to do as many questions as possible and you will soon see that VCE maths is just about perfecting what you already know.

kenhung123 · « **Reply #85 on:** December 06, 2009, 04:49:59 pm »

Do you solve the inverse function through Method 1 or 2?

TrueTears · « **Reply #86 on:** December 06, 2009, 04:54:37 pm »

You solve inverse functions by solving what they want you to solve.

kenhung123 · « **Reply #87 on:** December 06, 2009, 04:56:37 pm »

Do you do interchanging method (method 1) or f1^-1 method (Method 2)?

TrueTears · « **Reply #88 on:** December 06, 2009, 04:58:39 pm »

Oh well judging by your last question I thought you meant solve an inverse equation

Now I know you meant "How do I find an inverse function"

You can do any method that suits you. In maths there are many ways of doing one question.

kenhung123 · « **Reply #89 on:** December 06, 2009, 05:15:47 pm »

How do I inverse g(x)=x^2+2x by interchanging?

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