Question Regarding Function Graphs (1 Viewer)

AsuTeksu · Oct 13, 2023

Howdy All!

I have a question regarding function graphs as specified in the title above.

I've been looking at HSC Papers and Question 10 always seems to stump me ;-; and for that matter, any question that contains similar content.

Question 10 for 2020 can be seen below:

Screenshot 2023-10-13 at 10.01.12 am.png

Question 10 for 2022 can be seen below:

Screenshot 2023-10-13 at 10.01.45 am.png

When I see questions like this, I'm really unsure of how to answer them, or better yet, how to even approach them.

If anyone could help out and let me know how to solve them or a general approach I could adopt, it would be greatly appreciated!

Thanks in advance c:

SadCeliac · Oct 13, 2023

temporarylol said:
Stationary points would occur when h'(x) = 0, since h(x) = f(g(x)) then h'(x) = f'(g(x)) x g'(x) (using chain rule) so either we need to find points when g'(x) = 0 OR when f'(g(x)) = 0.

Clearly g'(x) = 0 occurs at x = 3 (ie g'(3) = 0) since there is a stationary point. Furthermore f'(x) = 0 occurs roughly at x = 1 (ie f'(1) = 0), but since g(x) needs to be the input, we need to find when g(x) = 1. When looking at the graph g(x) = 1 when x = 1.5 and x = 4.5.

So now we have three solutions that produce stationary points for h(x), the points being x = 1, x = 1.5 and x = 4.5.

I would just get a couple of points: in g(f(x)) the output of f(x) serves as the input of g(x). So let's see the point x = 1: f(1) = 2 (roughly) and since 2 is the output of f(x) this will now be the input for g(x), giving us g(2) = 1 (roughly). From this we get the result that h(1) = 1, which would eliminate C and D.

Just keep repeating this process to get a couple more points, but anyways I think the answer would be B since in A, you can see that h(-5) = -1 (roughly). However, that doesn't make sense since at x = -5: f(-5) = 25 (roughly) and g(25) = 1 (roughly), not -1.

@temporarylol why did you delete all of your messages

(except for the write a word starting with the last letter thread messages

)

AsuTeksu · Oct 13, 2023

temporarylol said:
Stationary points would occur when h'(x) = 0, since h(x) = f(g(x)) then h'(x) = f'(g(x)) x g'(x) (using chain rule) so either we need to find points when g'(x) = 0 OR when f'(g(x)) = 0.

Clearly g'(x) = 0 occurs at x = 3 (ie g'(3) = 0) since there is a stationary point. Furthermore f'(x) = 0 occurs roughly at x = 1 (ie f'(1) = 0), but since g(x) needs to be the input, we need to find when g(x) = 1. When looking at the graph g(x) = 1 when x = 1.5 and x = 4.5.

So now we have three solutions that produce stationary points for h(x), the points being x = 1, x = 1.5 and x = 4.5.

I would just get a couple of points: in g(f(x)) the output of f(x) serves as the input of g(x). So let's see the point x = 1: f(1) = 2 (roughly) and since 2 is the output of f(x) this will now be the input for g(x), giving us g(2) = 1 (roughly). From this we get the result that h(1) = 1, which would eliminate C and D.

Just keep repeating this process to get a couple more points, but anyways I think the answer would be B since in A, you can see that h(-5) = -1 (roughly). However, that doesn't make sense since at x = -5: f(-5) = 25 (roughly) and g(25) = 1 (roughly), not -1.

Thank you so much! Words cannot express how grateful I am! I hope you receive good luck for the next 100 years

Luukas.2 · Oct 13, 2023

View attachment 40447

Question 10 of the 2020 Advance Maths HSC was a challenging question.

There are (at least) two approaches that will work here. One is to use theory and differential calculus, which is quick but sophisticated. The other is to take a more straight-forward but longer approach.

Starting with the second of these:

$\begin{align*} \text{When $x = 0$, } g(x) &\approx 4.5 \quad \implies \quad h(0) = f(g(0)) \approx (f(4.5) \approx 0.2 \quad \implies \quad (0,\,0.2) \text{ lies on $y = h(x)$} \\ \text{When $x = 1$, } g(x) &\approx 3.5 \quad \implies \quad h(1) = f(g(1)) \approx (f(3.5) \approx 0.8 \quad \implies \quad (1,\,0.8) \text{ lies on $y = h(x)$} \\ \text{When $x = 2$, } g(x) &\approx 0.5 \quad \implies \quad h(2) = f(g(2)) \approx (f(0.5) \approx 1.8 \quad \implies \quad (2,\,1.8) \text{ lies on $y = h(x)$} \\ \text{When $x = 3$, } g(x) &= 0 \qquad \implies \quad h(3) = f(g(3)) = (f(0) \approx 1.7 \qquad \implies \quad (3,\,1.7) \text{ lies on $y = h(x)$} \\ \text{When $x = 4$, } g(x) &\approx 0.5 \quad \implies \quad h(4) = f(g(4)) \approx (f(0.5) \approx 1.8 \quad \implies \quad (4,\,1.8) \text{ lies on $y = h(x)$} \\ \text{When $x = 5$, } g(x) &\approx 3.5 \quad \implies \quad h(5) = f(g(5)) \approx (f(3.5) \approx 0.8 \quad \implies \quad (5,\,0.8) \text{ lies on $y = h(x)$} \end{align*}$

Plotting these points and adding a smooth curve shows that $y = h(x)$ increases from its y-intercept to a maximum somewhere near $x = 2$ , then decreases to a minimum at $x = 3$ ,and then rises again to a maximum somewhere near $x = 4$ , before decreasing again towards the x-axis, which is a horizontal asymptote. There are clearly 3 stationary points in the domain of interest.

The more theoretical approach is:

$\begin{align*} y &= h(x) = f(g(x)) \\ \frac{dy}{dx} &= f'(g(x)) \times g'(x) \qquad \text{using the Chain Rule} \\ \text{So, for stationary points:} \qquad f'(g(x)) &= 0 \qquad \text{or} \qquad g'(x) = 0 \end{align*}$

Consider $g'(x) = 0$ : The graph of $y = g(x)$ shows a parabola-like shape with a single stationary point at $x = 3$ . Thus, $y = h(x)$ will have a stationary point at $x = 3$ .

Consider $f'(g(x)) = 0$ : The graph of $y = f(x)$ shows a maximum turning point at about (1, 2), meaning that $f'(1) = 0$ , and so stationary points will occur whenever $g(x) \approx 1$ . The line $y = 1$ crosses the graph of $y = g(x)$ twice in the domain of interest, somewhere near $x = 2$ and $x = 4$ .

It follows that the graph of $y = h(x)$ will show three stationary points for $1 \le x \le 5$

AsuTeksu · Oct 13, 2023

@Luukas.2

Thank you so much! The time you took to do all that is appreciated

I hope life rewards you for eternity

Now I have no excuse to fail questions like this of any sort-

Question Regarding Function Graphs (1 Viewer)

AsuTeksu

Rising Renewal

SadCeliac

done hsc yay

AsuTeksu

Rising Renewal

Luukas.2

Well-Known Member

AsuTeksu

Rising Renewal

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