06 4sketchingcurves

Curve Sketching

Stationary Points

A stationary point is the place on a graph where the tangent to the curve is horizontal

• ie the gradient equals zero.

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The stationary points for any function can be found by making the derivative equal to zero.

• ie we want $\dfrac{dy}{dx} = 0$

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Types of Stationary Point

There are 3 types of stationary point

1. . Local Minimum
2. . Local Maximum
3. . Stationary Point of Inflection

• Local Minimums and Local Maximums are also called turning points.
  • The gradient "turns" from positive to negative (or negative to positive) on each side of the turning point.

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In the graph shown here:

• A is a local minimum
• B is a stationary point of inflection
• C is a local maximum

Note: The point A is not the absolute minimum value of the graph. It is only a minimum in the area local to the stationary point. In the same way, C is not the absolute maximum value of the graph.

Note: The point of inflection at B is the point on the graph where the gradient changes from concave down to concave up. In other words, the gradient was decreasing as it approached B and after B the gradient begins increasing.

Note: There are also two non-stationary points of inflection on this graph (not marked) where the concavity changes. One is between A and B, the other is between B and C. At each of these points the gradient stops increasing, achieves its steepest point and then starts decreasing.

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Summary

• A stationary point occurs where the gradient function has an x-intercept (ie derivative is zero).
  • A local minimum occurs where the gradient moves from negative to positive
  • A local maximum occurs where the gradient moves from positive to negative
• A stationary point of inflection occurs when the gradient function has a turning point on the x-axis.
• A non-stationary point of inflection occurs when the gradient function has a turning point that is not on the x-axis.

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Determining the Nature of a Stationary Point

• Once you have located the x-coordinate of a stationary point,
• the nature of the stationary point can be determined by constructing a gradient table.
• This involves finding the derivative (gradient) a small step on each side of the stationary point.

Example 1

Find the coordinates and nature of the stationary point in the function $y = x^2 - 4x + 1$
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Solution

… … Find the derivative:

… … $y = x^2 - 4x + 1$

… … $\dfrac{dy}{dx} = 2x - 4$
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… … Set the derivative to zero

… … $\dfrac{dy}{dx} = 2x - 4$

… … $2x - 4 = 0$

… … $2x = 4$

… … . $x = 2$
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… … Find y-coordinate

… … $y = x^2 - 4x + 1 \qquad substitute \; x = 2$

… … $y = (2)^2 - 4(2) + 1$

… … $y = 4 - 8 + 1$

… … $y = -3$
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… … Now construct gradient table for values close to $x = 2$

… … Inspection of the third row reveals that:

… … the stationary point is a local minimum

… … Coordinates of the local minimum are $(2, \; -3)$

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Example 2

Find the coordinates and nature of the stationary point in the function $f(x) = -x^3 +6x^2 - 12x + 11$

Solution

… … $f(x) = -x^3 + 6x^2 - 12x + 11$

… … $f'(x) = -3x^2 + 12x - 12$
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… … Let $f'(x) = 0$

… … $-3x^2 + 12x - 12 = 0$

… … $-3 (x - 2)^2 = 0$

… … … $x - 2 = 0$

… … … $x = 2$
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… … When $x = 2, \; y = 3$
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… … Gradient table for $x = 2$

… … Therefore $(2, \; 3)$ is a stationary point of inflection

• Notice that this stationary point of inflection occurs
• because the gradient function has a turning point located on the x-axis

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Locating Turning Points with your Calculator

Your Classpad calculator makes it easy to find the turning points on a graph.

• Graph the function in the Graphs-Tables section
• Make sure the window is sized so that the turning points are visible
• click on the Max or Min icon in the toolbar above the graph.
  • You may need to slide the toolbar to the right to see these two icons
• Max and Min commands are also in the Analysis Menu, G-Solve Submenu
• use the side arrows of your physical keypad to move between maximums (or minimums)

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OR

• Use the solve function in the Main section of the calculator
• Then you will need to determine the nature of each stationary point found

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Example 3

Find the maximum point in $y = \sin(x), \;\; x \in [0, \; \pi ]$

Solution

… … From the calculator,

… … The local maximum is found to be $(1.571, \; 1)$

… … To get the exact value, you would need to use Main:

… … In the Main section of the calculator:

… … Solve the equation formed by setting the derivative to zero

… … … $\dfrac{d}{dx} \Big( \sin(x) \Big) = 0$

… … … Note the calculator gives the general solution:

… … … $x = \pi n - \dfrac{\pi}{2} \; \text{ for } n \in Z$

… … … either select n = 1, for the desired value

… … … or solve using the restricted domain

… … Hence the local maximum is $\Big( \dfrac{\pi}{2}, \; 0 \Big)$
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Example 4

Sketch $y = 2x^3 - 5x^2 + 7$,
Label all axial intercepts and stationary points

Solution

… i) … Find the stationary points.

… … … $y = 2x^3 - 5x^2 + 7$

… … … $\dfrac{dy}{dx}=6x^2-10x$

… … … Let $\dfrac{dy}{dx}=0$

… … … $6x^2-10x=0$

… … … $2x(3x-5)=0$

… … … $x=0 \quad \textit{or} \quad x=\dfrac{5}{3}$
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… … … When $x = 0, \; y = 7$

… … … When $x=\dfrac{5}{3}, \; y=\dfrac{64}{27}$
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… ii) … Determine the nature of the stationary points.

… … … Gradient Table for $x = 0, \; \text{ and } \; x = \dfrac{5}{3}$

… … … Therefore, $(0, \; 7)$ is a local maximum turning point,

… … … and $\big( \dfrac{5}{3}, \; \dfrac{64}{27} \big)$ is a local minimum turning point.

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… iii) … Find axial intercepts:

… … … When $x = 0, \; y = 7$,
… … … the y-intercept is $(0, \; 7)$.

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… … … When $y = 0$,

… … … $2x^3 - 5x^2 + 7 = 0$

… … … $(x + 1)\big(2x^2 - 7x + 7\big) = 0$

… … … $x + 1 = 0 \qquad or \qquad 2x^2 - 7x + 7 = 0$

… … … $x = -1$ … … the quadratic term has no real solutions

… … … hence $(-1, \; 0)$ is the only x-intercept.
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… iv) … Sketch Graph

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