Sketching graphs C3 and C4

It's not explicitly stated anywhere what graphs I'm supposed to know how to sketch for these modules like it was for C1 (quadratic, cubic, circle) and then questions come up where you need to find the range of functions of forms1/x^3, x^1/2, x^4 and I have no idea how to sketch these so can't find the range

What I'm saying is does anyone have a list of the types of graphs you should be able to sketch for C3 and C4 (AQA) or just any tips about this in general?

You don't need to know how to sketch them. You just need to know the range and domain of basic functions like $e^x$, $\ln x$, $\sqrt{x}$, $\dfrac{1}{x}$, etc... then the domains and ranges of more complicated functions like $\ln(e^x + \frac{1}{x})$ can be determined by using these basic ones.

I'm familiar with basic functions, could you explain how you would get the range of a function like that

That function is a bit too much to ask an A-Level student (I've only used it as an illustration)

A simpler one would be to consider $\ln(x^2+4x+3)$.

The domain is straight forward since we require the input into the log function to be +ve hence $x^2+4x+3 > 0$ which yields that $\{ x < -3 \} \cup \{ x > -1 \}$.
Then for the range, observe that $x^2+4x+3$ goes between 0 and +infinity. For the log function, this means that it goes between $-\infty$ and $+\infty$, hence that's the range for this function.


Another example, $\sqrt{e^{x^2-4x+6}-1}$

The only restriction we got here is that the thing under the root is $\geq 0$ hence $e^{x^2-4x+6}-1 \geq 0 \implies x^2 - 4x+6 \geq 0 \implies x \in \mathbb{R}$. So that's the domain.
For the range, note that the square root function is an increasing function so minimising its input means minimising its value. To minimise the input, we say that $(2x-4)e^{x^2-4x+6} = 0 \implies x=2$ so the min of the function is given by $(2,\sqrt{e^2-1})$ (you can check that it's a min via the 2nd derivative). We don't have an upper bound so the range remains as $\sqrt{e^2-1} \leq \sqrt{e^{x^2-4x+6}-1} < \infty$

Doesn't it go between the minimum and infinity?

And I don't really understand the second part of the second example, what do you mean by minimising the input and minimising the value?

Could you work through finding the range for 2e^3x - 1 which was a question I got wrong on a past paper earlier today, it was already given in the question that the domain is all real values of x

Yeah, zero is its minimum (well, it doesn't actually obtain zero but it gets close) because we said $x^2+4x+3 > 0$



The smaller the number you plug into $\sqrt{x}$, the smaller the the result will be. So if we plug in the smallest thing we're allowed to, then we are going to obtain the smallest possible value of the square root function. Hence minimising the input minimised the output value.



This is one of the simpler ones than what I've shown. All you need to say here is that $0< e^x < \infty$ (for $x \in \mathbb{R}$) as a starting point, then manipulate all sides of the inequality in the same ways so that the middle becomes $2e^{3x}-1$

I understand the examples now

Okay so 0 < 2e^x < infinity
-1 < 2e^x - 1 < infinity

But I have no idea what happens when you multiply x by 3

Well, what does multiplying x by 3 do graphically? Does it translate/stretch the curve at all in the y-direction? So does that affect the range?

Erm it doesn't affect y so the range is just -1 to infinity?

Exactly.

Will the same method work with things like 1/x^4 ?

The method is more suitable to functions which have some sort of base to work off and apply a bunch of transformations to, i.e. for $2e^{3x}-1$ is obtained from the base function $e^x$ by applying bunch of linear transformations to.

Not so much for something like $\dfrac{1}{x^4}$, although no work is required to determine that clearly the range for it is between 0 and infinity.

The method is more suitable to functions which have some sort of base to work off and apply a bunch of transformations to, i.e. for $2e^{3x}-1$ is obtained from the base function $e^x$ by applying bunch of linear transformations to.

Not so much for something like $\dfrac{1}{x^4}$, although no work is required to determine that clearly the range for it is between 0 and infinity.