Curve-Sketching Using the Derivative (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Analysing Concavity and Finding Points of Inflection

For $f(x) = x^5 - 5x^4$:

a) Find any turning points

b) Draw up a table of concavities, find any points of inflection and the gradients of the inflectional tangents, describe the concavity, then sketch

Solution:

First, find the derivatives:

$f(x) = x^5 - 5x^4 = x^4(x - 5)$

$f'(x) = 5x^4 - 20x^3 = 5x^3(x - 4)$

$f''(x) = 20x^3 - 60x^2 = 20x^2(x - 3)$

Part a) Finding turning points

$f'(x)$ has zeroes at $x = 0$ and $x = 4$, with no discontinuities.

Create a table of test values for $f'(x)$:

The slope symbols show:

• / means increasing (positive gradient)
• \ means decreasing (negative gradient)
• — means zero gradient (stationary point)

From this table:

• $(0, 0)$ is a maximum turning point
• $(4, -256)$ is a minimum turning point

Part b) Analysing concavity

$f''(x)$ has zeroes at $x = 0$ and $x = 3$, with no discontinuities.

Create a table of test values for $f''(x)$:

The concavity symbols show:

• $\cap$ means concave down (negative $f''$)
• $\cup$ means concave up (positive $f''$)
• . indicates a potential point of inflection

From this analysis:

• $(3, -162)$ is a point of inflection (concavity changes from down to up)
• $(0, 0)$ is not a point of inflection (concavity remains down on both sides)

At the point of inflection $(3, -162)$:

$f'(3) = 5(3)^3(3 - 4) = 5(27)(-1) = -135$

So the inflectional tangent has gradient $-135$.

Concavity description:

• The graph is concave down for $x < 0$ and $0 < x < 3$
• The graph is concave up for $x > 3$

Worked Example: Identifying a Stationary Point of Inflection

Use the second derivative, if possible, to determine the nature of the stationary points of $f(x) = x^4 - 4x^3$. Find also any points of inflection, examine the concavity over the whole domain, and sketch the curve.

Solution:

Find the derivatives:

$f(x) = x^4 - 4x^3 = x^3(x - 4)$

$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$

$f''(x) = 12x^2 - 24x = 12x(x - 2)$

Finding stationary points:

$f'(x) = 0$ when $x = 0$ or $x = 3$

Test using the second derivative:

At $x = 3$: $f''(3) = 12(3)(3 - 2) = 12(3)(1) = 36 > 0$

Since $f''(3) > 0$, the point $(3, -27)$ is a minimum turning point.

At $x = 0$: $f''(0) = 0$

The second derivative test is inconclusive at $x = 0$. We need to check the first derivative around this point.

Create a table for $f'(x)$:

The function is decreasing on both sides of $x = 0$, so $(0, 0)$ is a stationary point of inflection.

Analysing concavity:

$f''(x) = 0$ when $x = 0$ or $x = 2$

Create a table for $f''(x)$:

From this table:

• $(0, 0)$ is the stationary point of inflection we already found (concavity changes from up to down)
• $(2, -16)$ is a non-stationary point of inflection (concavity changes from down to up)

At the point $(2, -16)$:

$f'(2) = 4(2)^2(2 - 3) = 4(4)(-1) = -16$

The inflectional tangent at $(2, -16)$ has gradient $-16$.

Concavity description:

• Concave up for $x < 0$ and $x > 2$
• Concave down for $0 < x < 2$

Worked Example: Determining Parameter Values for Given Concavity

For what values of $b$ is $y = x^4 - bx^3 + 5x^2 + 6x - 8$ concave down when $x = 2$?

Solution:

First, find the second derivative:

$y' = 4x^3 - 3bx^2 + 10x + 6$

$y'' = 12x^2 - 6bx + 10$

When $x = 2$:

$y'' = 12(2)^2 - 6b(2) + 10$

$y'' = 12(4) - 12b + 10$

$y'' = 48 - 12b + 10$

$y'' = 58 - 12b$

For the curve to be concave down at $x = 2$, we need $y'' < 0$:

$58 - 12b < 0$

$58 < 12b$

$\dfrac{58}{12} < b$

$4\dfrac{5}{6} < b$

Therefore, the curve is concave down at $x = 2$ when $b > 4\frac{5}{6}$.