Worked examples · Sheet 1

Relative maxima and minima with the first derivative

Theory used: First derivative test Where to look for extrema Relative maximum and minimum

Exercise 1

Find the relative maximum and minimum points of $f(x) = x^3 - 6x^2 + 9x + 1$ .

f'(x) = 3x^2 - 12x + 9 = 3\,(x^2 - 4x + 3) = 3\,(x - 1)(x - 3)

	$x < 1$	$1 < x < 3$	$x > 3$
$x - 1$	$-$	$+$	$+$
$x - 3$	$-$	$-$	$+$
$f'(x)$	$+$	$-$	$+$
$f$	increasing	decreasing	increasing

f(1) = 1 - 6 + 9 + 1 = 5, \qquad f(3) = 27 - 54 + 27 + 1 = 1

The point $x = 1$ is a relative maximum point, with $f(1) = 5$ ; the point $x = 3$ is a relative minimum point, with $f(3) = 1$ .

The function is a polynomial, so it is differentiable on all of $\mathbb{R}$ , and by Fermat’s theorem the relative extrema must be looked for among the stationary points. The derivative is a polynomial of degree two; taking out $3$ and factoring the trinomial $x^2 - 4x + 3$ , which has roots $1$ and $3$ , one obtains the product $3\,(x - 1)(x - 3)$ .

The sign of a product is obtained from the sign of the factors, as in the table. The derivative is positive before $1$ and negative between $1$ and $3$ , so by the first derivative test the point $x = 1$ is a relative maximum point. The derivative changes from negative to positive at $3$ , hence $x = 3$ is a relative minimum point. The values of the function are computed by substituting into the original polynomial. Neither of the two extrema is absolute, because a polynomial of degree three tends to $+\infty$ as $x \to +\infty$ and to $-\infty$ as $x \to -\infty$ .

Exercise 2

Find the relative maximum and minimum points of $f(x) = x^4 - 8x^2 + 3$ .

f'(x) = 4x^3 - 16x = 4x\,(x^2 - 4) = 4x\,(x - 2)(x + 2)

	$x < -2$	$-2 < x < 0$	$0 < x < 2$	$x > 2$
$4x$	$-$	$-$	$+$	$+$
$x^2 - 4$	$+$	$-$	$-$	$+$
$f'(x)$	$-$	$+$	$-$	$+$

f(-2) = 16 - 32 + 3 = -13, \qquad f(0) = 3, \qquad f(2) = -13

The points $x = \pm 2$ are relative minimum points, with value $-13$ ; the point $x = 0$ is a relative maximum point, with value $3$ .

The function is a polynomial, so it is differentiable everywhere and the only candidates are the stationary points. Taking out $4x$ in the derivative leaves the difference of squares $x^2 - 4$ , hence the derivative vanishes at $-2$ , $0$ and $2$ .

The factor $4x$ is positive for $x > 0$ , whereas $x^2 - 4$ is positive outside the interval $[-2, 2]$ . In the table the derivative changes sign from $-$ to $+$ at $-2$ and at $2$ , which by the first derivative test are relative minimum points, and from $+$ to $-$ at $0$ , which is a relative maximum point. The function contains only even powers of $x$ , so it is even, and indeed the two minima are symmetric and have the same value $-13$ . Since $f(x) \to +\infty$ as $x \to \pm\infty$ , the value $-13$ is also the absolute minimum.

Exercise 3

Find the relative maximum and minimum points of $f(x) = \sqrt[3]{x^2}\,(x - 5)$ .

f(x) = x^{\frac{5}{3}} - 5\,x^{\frac{2}{3}}, \qquad x \in \mathbb{R}

\begin{aligned} f'(x) &= \frac{5}{3}\,x^{\frac{2}{3}} - \frac{10}{3}\,x^{-\frac{1}{3}} \\[6pt] &= \frac{5x - 10}{3\,\sqrt[3]{x}}, \qquad x \neq 0 \end{aligned}

	$x < 0$	$0 < x < 2$	$x > 2$
$5x - 10$	$-$	$-$	$+$
$3\sqrt[3]{x}$	$-$	$+$	$+$
$f'(x)$	$+$	$-$	$+$

f(0) = 0, \qquad f(2) = \sqrt[3]{4}\,(2 - 5) = -3\sqrt[3]{4} \approx -4.76

The point $x = 0$ is a relative maximum point, with $f(0) = 0$ ; the point $x = 2$ is a relative minimum point, with $f(2) = -3\sqrt[3]{4}$ .

The cube root is defined for every real number, so the domain is $\mathbb{R}$ and the function is continuous. To differentiate, one writes the function as a sum of powers, multiplying $x^{\frac{2}{3}}$ by $x$ and by $-5$ . Bringing the derivative to the common denominator $3\,x^{\frac{1}{3}}$ one obtains a quotient that is not defined at $x = 0$ ; hence at $0$ the function is not differentiable, and $0$ must be added to the candidates together with the stationary point $x = 2$ .

The denominator $3\sqrt[3]{x}$ has the sign of $x$ , and the numerator is positive for $x > 2$ . The derivative is positive to the left of $0$ and negative to the right, so, since $f$ is continuous at $0$ , the first derivative test applies and $x = 0$ is a relative maximum point, even though at $0$ the function is not differentiable. At $x = 2$ the derivative changes from negative to positive, hence $x = 2$ is a relative minimum point. At the point $0$ the graph has a cusp, because $f'(x) \to +\infty$ from the left and $f'(x) \to -\infty$ from the right.