Worked examples · Sheet 2

Studies of functions with absolute value

Theory used: Scheme of curve sketching Relative maximum and minimum First derivative test Inflection point Qualitative graph

Go to: Exercise 1 Exercise 2

Exercise 1

Study the function

f (x) = x^{2} - 2∣ x ∣

and sketch its qualitative graph.

Domain. The expression is defined for every $x$ , therefore $D = R$ .

Symmetries and periodicity. $f (- x) = (- x)^{2} - 2∣ - x ∣ = x^{2} - 2∣ x ∣ = f (x)$ , therefore the function is even and the graph is symmetric with respect to the $y$ -axis; it is not periodic.

Sign and intersections. Since $x^{2} = ∣ x ∣^{2}$ , we can write $f (x) = ∣ x ∣ (∣ x ∣ - 2)$ . The factor $∣ x ∣$ is nonnegative, therefore $f (x) > 0$ for $∣ x ∣ > 2$ , $f (x) < 0$ for $0 < ∣ x ∣ < 2$ , and $f (x) = 0$ for $x = 0$ and $x = \pm 2$ .

Limits and asymptotes. The domain is unbounded on both sides and

x \to \pm \infty lim f (x) = + \infty.

Since $\frac{f ( x )}{x} = x - 2 \frac{∣ x ∣}{x} \to \pm \infty$ as $x \to \pm \infty$ , there are no oblique asymptotes; there are no vertical or horizontal asymptotes.

Continuity. $f$ is a sum of continuous functions, therefore it is continuous on $R$ .

Differentiability. Removing the absolute value,

f (x) = {x^{2} - 2 x x^{2} + 2 x if x \geq 0 if x < 0 f^{'} (x) = {2 x - 2 2 x + 2 if x > 0 if x < 0

At $0$ the one-sided limits of $f^{'}$ equal $- 2$ from the right and $2$ from the left; since they are finite and different, $0$ is a corner point and $f$ is differentiable on $R ∖ {0}$ .

Monotonicity, maxima and minima. For $x < 0$ we have $f^{'} (x) > 0$ when $x > - 1$ ; for $x > 0$ we have $f^{'} (x) > 0$ when $x > 1$ . By the monotonicity test $f$ is decreasing on $(- \infty, - 1]$ , increasing on $[- 1, 0]$ , decreasing on $[0, 1]$ , increasing on $[1, + \infty)$ . By the first derivative test $\pm 1$ are relative minimum points with $f (\pm 1) = - 1$ , while $0$ is a relative maximum point with $f (0) = 0$ , even without being stationary. Since $f (x) = (∣ x ∣ - 1)^{2} - 1 \geq - 1$ for every $x$ , the value $- 1$ is also the absolute minimum; the absolute maximum does not exist, because $f$ tends to $+ \infty$ .

Concavity and inflection points. On each of the two intervals $(- \infty, 0)$ and $(0, + \infty)$ we have $f^{''} (x) = 2 > 0$ , therefore $f$ is convex on each of them; there are no inflection points. On all of $R$ the function is not convex, because the segment joining the points $(- 1, - 1)$ and $(1, - 1)$ of the graph passes below the point $(0, 0)$ .

Graph.

y = x^{2} - 2 \left|x\right|

y = x^{2} - 2∣ x ∣

: corner point at

(0, 0)

, minima at

(- 1, - 1)

and

(1, - 1)

The delicate point is the origin. The function is continuous there, but the two expressions on either side have different slopes: the difference quotient tends to $2$ from the left and to $- 2$ from the right, therefore at $0$ the graph has a corner point. Fermat’s theorem does not apply, because it requires differentiability at the point; the relative maximum at $0$ is instead recognized with the first derivative test, which requires only continuity at $0$ and the sign of $f^{'}$ on the two sides, positive before and negative after.

The absolute minimum is justified without limits by completing the square: from $f (x) = ∣ x ∣^{2} - 2∣ x ∣ = (∣ x ∣ - 1)^{2} - 1$ it follows that $f (x) \geq - 1$ for every $x$ , and equality holds when $∣ x ∣ = 1$ . Comparing only the values at the candidates would not be enough, because the domain is not a closed and bounded interval and Weierstrass’s theorem does not apply.

Exercise 2

Study the function

f (x) = ∣ x^{2} - 1∣

and sketch its qualitative graph.

Domain. $D = R$ .

Symmetries and periodicity. $f (- x) = ∣ x^{2} - 1∣ = f (x)$ , therefore the function is even; it is not periodic.

Sign and intersections. An absolute value is always nonnegative, therefore $f (x) \geq 0$ for every $x$ and $f (x) = 0$ only where $x^{2} - 1 = 0$ , that is at $x = \pm 1$ . The intersection with the $y$ -axis is the point $(0, 1)$ .

Limits and asymptotes. $f (x) \to + \infty$ as $x \to \pm \infty$ ; since $\frac{f ( x )}{x} \to \pm \infty$ , there are no asymptotes of any kind.

Continuity. $f$ is the composition of the polynomial $x^{2} - 1$ with the absolute value, both continuous, therefore it is continuous on $R$ .

Differentiability. Removing the absolute value,

f (x) = {x^{2} - 1 1 - x^{2} if ∣ x ∣ \geq 1 if ∣ x ∣ < 1 f^{'} (x) = {2 x - 2 x if ∣ x ∣ > 1 if ∣ x ∣ < 1

At $x = 1$ the one-sided limits of $f^{'}$ equal $- 2$ from the left and $2$ from the right, therefore $1$ is a corner point; by symmetry so is $- 1$ . $f$ is differentiable on $R ∖ {- 1, 1}$ .

Monotonicity, maxima and minima. The sign of $f^{'}$ gives: $f$ decreasing on $(- \infty, - 1]$ , increasing on $[- 1, 0]$ , decreasing on $[0, 1]$ , increasing on $[1, + \infty)$ . By the first derivative test $0$ is a relative maximum point with $f (0) = 1$ , while $\pm 1$ are relative minimum points with $f (\pm 1) = 0$ . Since $f (x) \geq 0$ everywhere and the value $0$ is attained, $0$ is the absolute minimum; the absolute maximum does not exist.

Concavity and inflection points. For $∣ x ∣ > 1$ we have $f^{''} (x) = 2 > 0$ , therefore $f$ is convex on $(- \infty, - 1)$ and on $(1, + \infty)$ ; for $∣ x ∣ < 1$ we have $f^{''} (x) = - 2 < 0$ , therefore $f$ is concave on $(- 1, 1)$ . The concavity changes direction at $\pm 1$ , but at those points the graph does not have a tangent line, because they are corner points; therefore $\pm 1$ are not inflection points.

Graph.

y = \left|x^{2} - 1\right|

y = ∣ x^{2} - 1∣

: corner points at

(- 1, 0)

and

(1, 0)

, relative maximum at

(0, 1)

The graph is obtained from that of the parabola $y = x^{2} - 1$ by reflecting above the $x$ -axis the part between $- 1$ and $1$ , where the parabola is negative. The reflection explains both the corner points, produced by the abrupt change of slope at the points where the argument vanishes, and the relative maximum at $0$ , image of the vertex of the parabola.

At $\pm 1$ the concavity changes direction, but the definition of inflection point requires the existence of the tangent line at the point. Since at a corner point the two tangent half-lines are distinct, the tangent does not exist, hence $\pm 1$ are not inflection points; they are minimum points with a change of concavity.