Worked examples · Sheet 2

Concavity and inflection points

Theory used: Convex function and concave function Sign of the second derivative and concavity Inflection point Second derivative and inflections

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Find the intervals of convexity and concavity and the inflection points of $f (x) = x^{4} - 6 x^{2}$ , and write the equation of the tangent at the inflection with positive abscissa.

f^{'} (x) = 4 x^{3} - 12 x, f^{''} (x) = 12 x^{2} - 12 = 12 (x - 1) (x + 1)

	$x < - 1$	$- 1 < x < 1$	$x > 1$
$f^{''} (x)$	$+$	$-$	$+$
$f$	convex	concave	convex

f (\pm 1) = 1 - 6 = - 5, f^{'} (1) = 4 - 12 = - 8

y = - 5 - 8 (x - 1) = - 8 x + 3

The function is convex on $(- \infty, - 1]$ and on $[1, + \infty)$ , concave on $[- 1, 1]$ ; the inflections are $(- 1, - 5)$ and $(1, - 5)$ , and the tangent at the second one is $y = - 8 x + 3$ .

The function is a polynomial, so it is twice differentiable on $R$ and concavity is obtained from the sign of the second derivative. The second derivative $12 (x^{2} - 1)$ is a trinomial with roots $\pm 1$ and positive leading coefficient, hence it is positive outside the interval $[- 1, 1]$ and negative inside. By the test on the sign of the second derivative the function is convex where $f^{''} > 0$ and concave where $f^{''} < 0$ .

The second derivative changes sign at $- 1$ and at $1$ , so by the sufficient condition these two points are inflections. The inflection tangent is written with the formula of the tangent line, using the value $f (1) = - 5$ and the slope $f^{'} (1) = - 8$ . Since $f^{'} (1) \neq = 0$ , the inflection is oblique.

Exercise 2

Find the intervals of convexity and concavity and the inflection points of $f (x) = x e^{x}$ .

f^{'} (x) = e^{x} + x e^{x} = (x + 1) e^{x}

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

f^{''} (x) > 0 ⟺ x + 2 > 0 ⟺ x > - 2

f (- 2) = - 2 e^{- 2} = - \frac{2}{e ^{2}} \approx - 0.27

The function is concave on $(- \infty, - 2]$ and convex on $[- 2, + \infty)$ , with an inflection at the point $(- 2, - \frac{2}{e ^{2}})$ .

The function is defined and infinitely differentiable on $R$ , because it is the product of a polynomial and an exponential. The first derivative is computed with the product rule and $e^{x}$ is taken out; the second derivative is obtained by differentiating the product $(x + 1) e^{x}$ again.

The factor $e^{x}$ is always positive, so the sign of $f^{''}$ is that of $x + 2$ . The second derivative is negative for $x < - 2$ and positive for $x > - 2$ , hence the function is concave to the left of $- 2$ and convex to the right. The change of sign of $f^{''}$ at $- 2$ indicates that $x = - 2$ is an inflection point, and its value is obtained by substituting into the function.

Exercise 3

Find the intervals of convexity and concavity and the inflection points of $f (x) = ln (1 + x^{2})$ , and write the equation of the tangent at the inflection with positive abscissa.

f^{'} (x) = \frac{2 x}{1 + x ^{2}}

f^{''} (x) = \frac{2 ( 1 + x ^{2} ) - 2 x \cdot 2 x}{( 1 + x ^{2} ) ^{2}} = \frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}}

f^{''} (x) > 0 ⟺ 1 - x^{2} > 0 ⟺ - 1 < x < 1

f (\pm 1) = ln 2, f^{'} (1) = 1, y = ln 2 + (x - 1)

The function is convex on $[- 1, 1]$ and concave on $(- \infty, - 1]$ and on $[1, + \infty)$ ; the inflections are $(\pm 1, ln 2)$ , and the tangent at the inflection with abscissa $1$ is $y = x - 1 + ln 2$ .

y = \ln(1 + x^2)

y = x - 1 + \ln 2

The function

ln (1 + x^{2})

is convex between the inflections

(\pm 1, ln 2)

and concave outside; at the inflection

(1, ln 2)

the graph crosses the tangent

The argument of the logarithm $1 + x^{2}$ is always positive, so the domain is $R$ . The first derivative is obtained with the chain rule, and the second derivative with the quotient rule; in the numerator the terms $2 x^{2}$ and $- 4 x^{2}$ add up to $- 2 x^{2}$ , and $2$ is taken out.

The denominator $(1 + x^{2})^{2}$ is positive, hence the sign of $f^{''}$ is that of $1 - x^{2}$ , positive for $- 1 < x < 1$ and negative outside. The function is therefore convex on the interval $[- 1, 1]$ and concave outside, and the points $\pm 1$ , where $f^{''}$ changes sign, are inflections. The function is even, so the two inflections are symmetric with respect to the $y$ -axis. The tangent at the inflection with abscissa $1$ has slope $f^{'} (1) = \frac{2}{2} = 1$ .