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VCE Methods Question Thread!
1729:
--- Quote from: PizzaMaster on June 27, 2022, 03:32:27 pm ---Thank you for your reply!
however, is it possible to explain to me assuming I have no knowledge of differentiation
--- End quote ---
You need to have some knowledge of differentiation to determine whether a cubic has 0, 1 or 2 turning points.
Just adding to Billuminati's response, if you want to know how many turning points a quadratic has then you can just take the discriminant of the derivative. \(f\left(x\right)=ax^3+bx^2+cx+d\) then \(f'\left(x\right)=3ax^{2}+2bx+c\)
The discriminant of the derivative is equal to, \(\Delta _{f'}=4b^2-12ac\).
\(\Delta _{f'}<0\Longrightarrow \text{0 stationary points}\)
\(\Delta _{f'}>0\Longrightarrow \text{2 stationary points}\)
\(\Delta _{f'}=0\Longrightarrow \text{1 stationary point}\)
Note: 2 stationary points implies that the stationary points are turning points, and 1 stationary point means that the stationary point is an inflection point.
So I guess what you could do, without knowledge of differentiation, is just plug in \(a,\:b,\:c\) for a cubic in the form \(ax^3+bx^2+cx+d\) into the expression \(4b^2-12ac\) to determine the amount of turning points.
This is a little formulaic, which I don't really like but I guess if you do not know differentiation, just use this.
beep boop :
Hi all,
Can someone please help me w/ q17 part g?
Much appreciated,
beep boop
1729:
--- Quote from: beep boop on June 27, 2022, 06:24:35 pm ---Hi all,
Can someone please help me w/ q17 part g?
Much appreciated,
beep boop
--- End quote ---
Hi beepboop
Assuming you were able to correctly answer the previous parts of the question.
Part F.
Consider how \(f^{-1}\) is just a reflection over \(y=x\), the shaded regions essentially remain the same, however, the location of the \(x\) coordinate of the point of tangency is different.
If the shaded regions \(f\) makes with it's tangent are at a minimum when the point of tangency is at \(x=10^{-\frac{1}{3}}\), then the shaded regions \(f^{-1}\) makes with it's tangent is at a minimum when the point of tangency is at \(f\left(10^{-\frac{1}{3}}\right)\).
This is because a reflection in \(y=x\) maps the coordinates of \(f\) as such \(\left(a,f\left(a\right)\right)\rightarrow \left(f\left(a\right),a\right)\)
Part G.
The tangent of \(f\) at \(x=1\) would be \(f'\left(1\right)\left(x-1\right)+f\left(1\right)=3-3x\).
However, \(f^{-1}\) is not actually differentiable at \(x=1\), therefore, having a vertical tangent of \(x=1\). We know that the angle formed with the \(x\) axis and the tangent of \(f\) in the clock-wise negative direction is \(\left|\tan^{-1}\left(-3\right)\right|=\tan^{-1}\left(3\right)\) and the angle formed with \(x\) axis in clockwise negative direction is \(\frac{\pi}{2}\) as it's vertical.
Therefore the acute angle formed is \(\frac{\pi}{2}-\tan^{-1}\left(3\right)\).
The graph below may help visualising this, I essentially took the approach of subtracting red angle with green.
PizzaMaster:
Guyss
I don't get it
is square root 16
4
or
+- 4
Also,
when we have 9^1/2
is that 3 or +- 3
Thank you!
1729:
--- Quote from: PizzaMaster on June 28, 2022, 04:28:12 pm ---Guyss
I don't get it
is square root 16
4
or
+- 4
Also,
when we have 9^1/2
is that 3 or +- 3
Thank you!
--- End quote ---
Hi, this is actually a very good question which many students fail to get right.
\(x^2=16\) has two solutions, \(x=4\) and \(x=-4\)
But \(x=\sqrt{16}\) has only one solution, \(x=4\)
To help understand why this is the case, try and consider it visually... the graph of \(y=\sqrt{x}\) is strictly positive (or 0)
Pretty much the square root of a number by itself is always positive. Also, \(x^{\frac{1}{2}}=\sqrt{x}\).
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