# Thread: A level maths 06

1. ## A level maths 06

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

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

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

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

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

$\displaystyle \Rightarrow 5 = 2 + a$

$\displaystyle \Rightarrow a = 3$

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

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

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

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

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

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

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

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

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

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

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

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

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