A level maths 06

**carlasader** · May 24th 2007, 01:02 PM

some more ...i have so many things to solve but i dont know where to start from ,so urgent help needed!!!

**Jhevon** · May 24th 2007, 01:50 PM

Originally Posted by carlasader

some more ...i have so many things to solve but i dont know where to start from ,so urgent help needed!!!

us drawing the diagram for you won't help, since we'd use a graphing utility and you won't know what happens. if you have any specific questions about graphing, just say so and i'll try to answer them. otherwise i'll find a from question 2

$y = (x - 1)^2 (x + a)$

Now we are told that the graph passes through (2,5), this means when x = 2, y = 5

$\Rightarrow 5 = (2 - 1)^2 (2 + a)$

$\Rightarrow 5 = 2 + a$

$\Rightarrow a = 3$

Hopefully you can take it from here. if not, ask SPECIFIC questions about graphing

**Jhevon** · May 24th 2007, 02:04 PM

Originally Posted by carlasader

some more ...i have so many things to solve but i dont know where to start from ,so urgent help needed!!!

here's 3

Let $h(x) = x^3$
Let $f(x) = x - 1$
Let $g(x) = 2x$

Then we can form $y = 2(x - 1)^3$ via a composite function of f, g and h

Let $y = (g \circ h \circ f )(x)$

$\Rightarrow y = g(h(f(x)))$

$\Rightarrow y = g((x - 1)^3)$

$\Rightarrow y = 2(x - 1)^3$

As i said, graphing for you won't help, so let's move on to finding the inverse of y

$y = 2(x - 1)^3$

For inverse, we solve $x = 2(y - 1)^3$ for $y$

$\Rightarrow \frac {x}{2} = (y - 1)^3$

$\Rightarrow \sqrt [3] { \frac {x}{2}} = y - 1$

$\Rightarrow y = \sqrt [3] { \frac {x}{2}} - 1$ is the inverse function

As for the graphical relationship. we obtain the graph of the inverse of a function by reflecting the original function in the line $y = x$

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