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July 15, 2020, 02:29:24 pm

Author Topic: Curve sketching using differentiation  (Read 401 times)

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Curve sketching using differentiation
« on: June 10, 2019, 10:45:37 am »
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A common type of question they'll ask looks like this:
\[ \text{Sketch } y=x^3-3x^2-24x+9\text{ stating all stationary points and inflection points.}\]
The examiners are usually not interested in your ability to track down the \(x\)-intercepts. This is because the expressions given are usually nasty to factorise. The idea is that you're able to get a strong idea of what the curve should look like, provided you only use stationary points and inflection points! (And maybe the \(y\)-intercept as well.)

Step 0: \(y\)-intercepts
If we're ever told to find the \(y\)-intercept, we should. This is done by subbing in \(x=0\) to obtain \(\boxed{y=9}\).

Step 1: Tracking down stationary points
Recall that stationary points are found by setting the first derivative to \(0\).
\[ \frac{dy}{dx} =3x^2-6x-24 \]
Setting this to \(0\) gives
0 &= 3x^2-6x-24\\
&= 3(x^2-2x-8)\\
&= 3(x-4)(x+2)\\
\therefore x &= -2, 4
Make sure that you find the corresponding \(y\)-coordinates, or else how will you plot them!
\[ \text{When }x=-2,\, y= 37\\ \text{When }x=4, \, y=-71\]
Step 2: Test for nature
- Approach 1: In questions where the examiners know that the second derivative is a catastrophe to compute, they will not make you find points of inflection. Instead, it is better to test values on both sides of each stationary point.

For this question, obviously this isn't necessary, but we demonstrate the idea:
\[ \text{To the left of }x=-2: \text{ test, say, }x=-3.\\ \text{At }x=-3, \, \frac{dy}{dx} =21 >0 \]
\[ \text{To the right of }x=-2\text{ and left of }x=4:\text{ test, say, }x=0.\\ \text{At }x=0,\, \frac{dy}{dx} = -24 <0 \]
\[ \text{To the right of }x=4:\text{ test, say, }x=5.\\ \text{At } x=5, \, \frac{dy}{dx} = 21 >0\]
So about \(x=-2\), the curve increases to \(x=-2\), then decreases. We deduce that \( (2,37)\) is a local maximum.

About \(x=4\), the curve decreases to \(x=4\), then increases. We deduce that \( (4, -71)\) is a local minimum.

- Approach 2: The concavity at the stationary point is enough to check its nature. For that we require the second derivative. This is recommended when the second derivative is easy to compute.
\[ \frac{d^2y}{dx^2}= 6x-6 \]
\[ \text{At }x=-2, \, \frac{d^2y}{dx^2} = -18 < 0\\ \text{At } x=2, \, \frac{d^2y}{dx^2} = 18 > 0 \]
So at \(x=-2\) the curve is concave up, and \( (2,37)\) is a local maximum. At \(x=4\), the curve is concave down, so \( (4,-71)\) is a local minimum.

(Note: This method is a tad annoying when we find \( \frac{d^2y}{dx^2} = 0\). When that happens, we need to test for a concavity change as in step 3, before we can conclude its nature.)

Step 3: State all candidate points of inflection
Recall that inflection points set the second derivative to \(0\), but not every point with second derivative \(0\) gives an inflection point. We therefore find all possible points of inflection first.
\frac{d^2y}{dx^2} &= 0\\

Step 4: Test for concavity change
Unlike with stationary points, we must always test for a change in concavity here. To be lazy, I'll just recycle values I've found above.
\[ \text{To the left of }x=1,\, \text{ test }x=-2:\\ \text{When }x=-2, \,\frac{d^2y}{dx^2} = -18 < 0 \]
\[ \text{To the right of }x=1, \, \text{ test } x=4:\\ \text{When }x=4,\, \frac{d^2y}{dx^2}=18 > 0\]
We see the concavity does change, so an inflection point does occur at \(x=2\). The \(y\)-coordinate is \(-17\).

Step 5: Sketch the curve!
On your sketch, you probably should label all the points you found clearly, instead of be lazy and use Desmos like I did.
You're going to be seeing this type of problem very often so make sure you know the method off by heart!