Worked examples · Sheet 3

Studies of functions with exponentials

Theory used: Scheme of curve sketching Monotonicity test First derivative test Necessary condition for an inflection point Absolute maximum and minimum Qualitative graph

Go to: Exercise 1 Exercise 2

Exercise 1

Study the function

f (x) = x e^{- x}

and sketch its qualitative graph.

Domain. The exponential is defined everywhere, therefore $D = R$ .

Symmetries and periodicity. $f (- x) = - x e^{x}$ is different from both $f (x)$ and $- f (x)$ , therefore the function is neither even nor odd; it is not periodic.

Sign and intersections. The factor $e^{- x}$ is always positive, therefore the sign of $f$ is the sign of $x$ : $f (x) > 0$ for $x > 0$ , $f (x) < 0$ for $x < 0$ , $f (0) = 0$ . The graph meets the axes only at the origin.

Limits and asymptotes. As $x \to + \infty$ the ratio $\frac{x}{e ^{x}}$ tends to $0$ by the hierarchy of infinities, therefore

x \to + \infty lim x e^{- x} = 0^{+}, x \to - \infty lim x e^{- x} = - \infty.

The line $y = 0$ is a horizontal asymptote as $x \to + \infty$ . As $x \to - \infty$ we have $\frac{f ( x )}{x} = e^{- x} \to + \infty$ , therefore there is no oblique asymptote; there are no vertical asymptotes.

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

Differentiability and monotonicity. $f$ is differentiable on $R$ and

f^{'} (x) = e^{- x} - x e^{- x} = (1 - x) e^{- x} .

Since $e^{- x} > 0$ , we have $f^{'} (x) > 0$ for $x < 1$ . By the monotonicity test $f$ is increasing on $(- \infty, 1]$ and decreasing on $[1, + \infty)$ .

Maxima and minima. The only stationary point is $1$ , where the derivative changes from positive to negative, therefore by the first derivative test $1$ is a relative maximum point with $f (1) = \frac{1}{e}$ . Since $f$ increases before $1$ and decreases after, the value $\frac{1}{e}$ is also the absolute maximum. The absolute minimum does not exist, because $f \to - \infty$ as $x \to - \infty$ .

Concavity and inflection points. Differentiating again,

f^{''} (x) = - e^{- x} - (1 - x) e^{- x} = (x - 2) e^{- x},

therefore $f^{''} (x) > 0$ for $x > 2$ . The function is concave on $(- \infty, 2]$ and convex on $[2, + \infty)$ ; since $f^{''}$ changes sign at $2$ , the point $2$ is an inflection point, with $f (2) = \frac{2}{e ^{2}}$ .

Graph.

y = x e^{-x}

y = x e^{- x}

: maximum at

(1, \frac{1}{e})

, inflection point at

(2, \frac{2}{e ^{2}})

, horizontal asymptote

y = 0

x \to + \infty

The limit as $x \to + \infty$ is a form of the type $\infty \cdot 0$ , which is resolved by rewriting the product as the ratio $\frac{x}{e ^{x}}$ . By the hierarchy of infinities the exponential grows faster than every power, therefore the ratio tends to $0$ ; the function stays positive for $x > 0$ , hence the graph approaches the asymptote $y = 0$ from above.

The relative maximum at $1$ is also absolute because of the pattern of monotonicity, since the function increases on all of $(- \infty, 1]$ and decreases on all of $[1, + \infty)$ : every other value of $f$ is less than $f (1)$ . The inflection point at $2$ is consistent with the horizontal asymptote, since a concave function cannot flatten onto a horizontal line while staying above it, and the change of concavity occurs just after the maximum.

Exercise 2

Study the function

f (x) = e^{- x^{2}}

and sketch its qualitative graph.

Domain. $D = R$ .

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

Sign and intersections. An exponential is always positive, therefore $f (x) > 0$ for every $x$ and the graph does not meet the $x$ -axis; it meets the $y$ -axis at the point $(0, 1)$ .

Limits and asymptotes. As $x \to \pm \infty$ the exponent $- x^{2}$ tends to $- \infty$ , therefore

x \to \pm \infty lim e^{- x^{2}} = 0^{+}

and the line $y = 0$ is a horizontal asymptote on both sides; there are no vertical asymptotes.

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

Differentiability and monotonicity. $f$ is differentiable on $R$ and

f^{'} (x) = - 2 x e^{- x^{2}} .

Since $e^{- x^{2}} > 0$ , we have $f^{'} (x) > 0$ for $x < 0$ . By the monotonicity test $f$ is increasing on $(- \infty, 0]$ and decreasing on $[0, + \infty)$ .

Maxima and minima. The only stationary point is $0$ , where the derivative changes from positive to negative, therefore $0$ is a relative maximum point and, because of the pattern of monotonicity, also an absolute one, with $f (0) = 1$ . The absolute minimum does not exist: $f$ is positive and tends to $0$ , therefore $0$ is the infimum but it is not an attained value.

Concavity and inflection points. Differentiating the product,

f^{''} (x) = - 2 e^{- x^{2}} + 4 x^{2} e^{- x^{2}} = (4 x^{2} - 2) e^{- x^{2}},

therefore $f^{''} (x) > 0$ for $x^{2} > \frac{1}{2}$ , that is for $∣ x ∣ > \frac{2}{2}$ . The function is convex on $(- \infty, - \frac{2}{2}]$ , concave on $[- \frac{2}{2}, \frac{2}{2}]$ , convex on $[\frac{2}{2}, + \infty)$ . At the points $\pm \frac{2}{2}$ the second derivative changes sign, therefore both are inflection points, with $f (\pm \frac{2}{2}) = e^{- \frac{1}{2}} = \frac{1}{e}$ .

Graph.

y = e^{-x^{2}}

y = e^{- x^{2}}

: maximum at

(0, 1)

, inflection points at

x = \pm \frac{2}{2}

, horizontal asymptote

y = 0

The evenness, verified at the beginning, organizes the whole study: it suffices to derive monotonicity, concavity and notable points for $x \geq 0$ and reflect the result across the $y$ -axis. In particular the two inflection points are symmetric and have the same ordinate.

The distinction between infimum and minimum is the step to justify carefully. The function is always positive and tends to $0$ at infinity, therefore no positive value is a better lower bound than $0$ ; on the other hand $f (x) = 0$ has no solutions, hence the infimum $0$ is not attained and the absolute minimum does not exist. Weierstrass’s theorem does not contradict this fact, because the domain $R$ is not a bounded interval and the hypothesis of the theorem is not satisfied.