Tag: Zeros

Use derivatives to sketch the graph of a function

by RoRi

September 29, 2015

Let

$f(x) = x^3 - 6x^2 + 9x + 5.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = 3x^2 - 12x + 9.$

Thus,

$f'(x) = 0 \quad \implies \quad 3x^2 - 12x + 9 = 0 \quad \implies \quad x = \{ 1, 3 \}.$
$f$ is increasing for $x < 1$ and $x > 3$ , and decreasing for $1 < x < 3$ .
Taking the second derivative,
$f''(x) = 6x - 12.$

Thus, $f'$ is increasing if $x > 2$ , and decreasing if $x < 2$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 28, 2015

Let

$f(x) = (x-1)^2 (x+2).$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$\begin{align*} f'(x) &= 2(x-1)(x+2) + (x-1)^2 \\ &= (x-1)(2x + 4 + x - 1) \\ &= (x-1)(3x + 3) \\ &= 3(x^2 -1) \end{align*}$

Thus,

$f'(x) = 0 \quad \implies \quad 3(x^2-1) = 0 \quad \implies \quad x = \pm 1.$
$f$ is increasing for $|x| > 1$ , and decreasing for $|x| < 1$ .
Taking the second derivative,
$f'(x) = 3(x^2-1) \quad \implies \quad f''(x) = 6x.$

Thus, $f'$ is increasing if $x > 0$ , and decreasing if $x < 0$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 28, 2015

Let

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

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f(x) = x^3 - 4x \quad \implies \quad f'(x) = 3x^2 - 4.$

Thus,

$f'(x) = 0 \quad \implies \quad 3x^2 - 4 = 0 \quad \implies \quad x = \pm \frac{2}{\sqrt{3}}.$
Since $f'(x) > 0$ when $|x| > \frac{2}{\sqrt{3}}$ and $f'(x) < 0$ when $|x| < \frac{2}{\sqrt{3}}$ we have $f$ is increasing if $|x| > \frac{2}{\sqrt{3}}$ and $f$ is decreasing if $|x| < \frac{2}{\sqrt{3}}$ .
Taking the second derivative,
$f'(x) = 3x^2 - 4 \quad \implies \quad f''(x) = 6x.$

Thus, $f'$ is increasing for $x > 0$ ; and decreasing for $x < 0$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 28, 2015

Let

$f(x) = x^2 - 3x + 2.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f(x) = x^2 - 3x + 2 \quad \implies \quad f'(x) = 2x - 3.$

Thus,

$f'(x) = 0 \quad \implies \quad 2x - 3 = 0 \quad \implies \quad x = \frac{3}{2}.$
Since $f'(x) < 0$ when $x < \frac{3}{2}$ and $f'(x) > 0$ when $x > \frac{3}{2}$ we have $f$ is decreasing if $x < \frac{3}{2}$ and $f$ is increasing if $x > \frac{3}{2}$ .
Taking the second derivative,
$f'(x) = 2x - 3 \quad \implies \quad f''(x) = 2.$

Since this is positive for all $x$ , this means $f'$ is increasing for all $x$ .
We sketch the curve,

Prove the second derivative of a function with a given property must have a zero

by RoRi

September 28, 2015

Consider a function $f$ which is continuous everywhere on an interval $[a,b]$ and has a second derivative $f''$ everywhere on the open interval $(a,b)$ . Assume the chord joining two points $(a, f(a))$ and $(b, f(b))$ on the graph of the function intersects the graph of the function at a point $(c, f(c))$ with $c \in (a,b)$ . Prove there exists a point $t \in (a,b)$ such that $f''(t) = 0$ .

Proof. Let $g(x)$ be the equation of the line joining $(a, f(a))$ and $(b, f(b))$ . Then define a function

$h(x) = f(x) - g(x).$

Since $f$ and $g$ intersect at the values $a,b$ , and $c$ this means

$f(a) = g(a), \qquad f(b) = g(b), \qquad f(c) = g(c).$

By our definition of $h$ then we have

$h(a) = h(b) = h(c) = 0.$

Further, since $f'$ and $g'$ are continuous and differentiable on $(a,b)$ , we apply Rolle’s theorem twice: first on the interval $[a,c]$ and then on the interval $[c,b]$ . These two applications of Rolle’s theorem tells us there exist points $c_1$ and $c_2$ such that

$h'(c_1) = h'(c_2) = 0, \qquad \text{with } a < c_1 < c < c_2 < b.$

Then, we apply Rolle’s theorem for a third time, this time to the function $h'$ on the interval $[c_1, c_2]$ to conclude that there exists a $t \in (c_1, c_2)$ such that $h''(t) = 0$ . Then, since $t \in (c_1, c_2)$ we know $t \in (a,b)$ (since $a < c_1 < c_2 < b$ ). So,

$\begin{align*} h(x) = f(x) - g(x) && \implies && h'(x) &= f'(x) - g'(x) \\ && \implies && h''(x) &= f''(x) &(g''(x) = 0 \text{ since } g \text{ is linear}). \end{align*}$

Thus, $f''(t) = 0$ for some $t \in (a,b). \qquad \blacksquare$

Prove properties about the zeros of a polynomial and its derivatives

by RoRi

September 28, 2015

Consider a polynomial $f$ . We say a number $\alpha$ is a zero of multiplicity $m$ if

$f(x) = (x - \alpha)^m g(x),$

where $g(\alpha) \neq 0$ .

Prove that if the polynomial $f$ has $r$ zeros in $[a,b]$ , then its derivative $f'$ has at least $r-1$ zeros in $[a,b]$ . More generally, prove that the $k$ th derivative, $f^{(k)}$ has at least $r-k$ zeros in the interval.
Assume the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in the interval $[a,b]$ . What can we say about the number of zeros of $f$ in the interval?

Proof. Let $\alpha_1, \ldots, \alpha_k$ denote the $k$ distinct zeros of $f$ in $[a,b]$ and $m_1, \ldots, m_k$ their multiplicities, respectively. Thus, the total number of zeros is given by,
$r = \sum_{i=1}^k m_i.$

By the definition given in the problem, if $\alpha_i$ is a zero of $f$ of multiplicity $m_i$ then

$f(x) = (x - \alpha_i)^{m_i} g(x) \qquad \text{where } g(x) \neq 0.$

Taking the derivative (using the product rule), we have

$\begin{align*} f'(x) &= m_i (x - \alpha_i)^{m_i - 1} g(x) + (x- \alpha_i)^{m_i} g'(x)\\ &= (x - \alpha_i)^{m_i - 1} (m_i g(x) + (x - \alpha_i) g'(x)) \end{align*}$

Thus, again using the definition given in the problem, $\alpha_i$ is a zero of $f'(x)$ of multiplicity $m_i -1$ .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros $\alpha_i$ and $\alpha_j$ of $f$ there exists a number $c \in [\alpha_i, \alpha_j]$ (assuming, without loss of generality, that $\alpha_i < \alpha_j$ ) such that $f'(c) = 0$ . Hence, if $f$ has $k$ distinct zeros, then the mean value theorem guarantees $k-1$ numbers $c$ such that $f'(c) = 0$ . Thus, $f'$ has at least:

$\left(\sum_{i=1}^k (m_i -1)\right) + k - 1 = \left( \sum_{i=1}^k m_i \right) - 1 = r-1 \ \ \text{zeros}.$

By induction then, the $k$ th derivative $f^{(k)}(x)$ has at least $r-k$ zeros.
If the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in $[a,b]$ , then we can conclude that $f$ has at most $r+k$ zeros in $[a,b]$ .

Show that x^2 = x*sin x + cos x for exactly two real numbers x

by RoRi

September 27, 2015

Consider the equation

$x^2 = x \sin x + \cos x.$

Show that there are two values of $x \in \mathbb{R}$ such that the equation is satisfied.

Proof. Let $f(x) = x^2 - x \sin x - \cos x$ . (We want then to find the zeros of this function since these will be the points that $x^2 = x \sin + \cos x$ .) Then,

$f'(x) = 2x - \sin x - x \cos x + \sin x = x(2 - \cos x).$

Since $2 - \cos x \neq 0$ for any $x$ (since $\cos x \leq 1$ ), we have

$f'(x) = 0 \quad \iff \quad x = 0.$

Then, $f$ is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know $f$ has at most two zeros (if there were three or more, say $x_1, x_2,$ and $x_3$ , then there must be distinct numbers $c_1$ and $c_2$ with $x_1 < c_1 < x_2 < c_2 < x_3$ such that $f'(c_1) = f'(c_2) = 0$ , but we know there is only one $c$ such that $f'(c) = 0$ ).
Furthermore, $f$ has at least two zeros since $f(-\pi) = \pi^2 + 1 > 0$ , $f(0) = -1 < 0$ , and $f(\pi) = \pi^2 + 1 > 0$ . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of $f$ is at most two and at least 2. Hence, the number of zeros must be exactly two $. \qquad \blacksquare$

Explain why the absence of a zero does not violate Rolle’s theorem

by RoRi

September 27, 2015

Consider the function

$f(x) = 1 - x^{\frac{2}{3}}.$

Show that $f(1) = 0$ and $f(-1) = 0$ , but that the derivative $f'(x) \neq 0$ for all $x \in [-1,1]$ . Explain why this does not violate Rolle’s theorem.

Proof. First, we show that $f(1) = 0$ and $f(-1) = 0$ by a direct computation:

$\begin{align*} f(1) = 1 - (1)^{\frac{2}{3}} = 1 - 1 &= 0, \\ f(-1) = 1 - (-1)^{\frac{2}{3}} = 1 - (1^2)^\frac{1}{3} = 1 - 1 &= 0. \end{align*}$

Then, we compute the derivative,

$f(x) = 1 - x^{\frac{2}{3}} \quad \implies \quad f'(x) = -\frac{2}{3} x^{-\frac{1}{3}}.$

To show $f'(x) \neq 0$ for any $x \in [-1,1]$ we consider three cases:

If $x < 0$ then $x^{-\frac{1}{3}} < 0$ implies $f'(x) > 0$ (since $-\frac{2}{3}$ times a negative is positive).
If $x > 0$ then $x^{-\frac{1}{3}} > 0$ implies $f'(x) < 0$ (since $-\frac{2}{3}$ times a positive is then negative).
If $x = 0$ , then $f'(x)$ is undefined (since $x^{-\frac{1}{3}} = \frac{1}{x^{1/3}}$ ).

Thus, $f'(x) \neq 0$ for any $x \in [-1,1]. \qquad \blacksquare$

This is not a violation of Rolle’s theorem since the theorem requires that $f(x)$ be differentiable for all $x$ on the open interval $(-1,1)$ . Since $f'(x)$ is not defined at $x = 0$ , we have $f(x)$ is not differentiable on the whole interval. Hence, Rolle’s theorem does not apply.

Prove there is only one zero of x³ – 3x + b in [-1,1]

by RoRi

September 26, 2015

Prove, using Rolle’s theorem, that for any value of $b$ there is at most one $x \in [-1,1]$ such that

$x^3 - 3x + b = 0.$

Proof. The argument is by contradiction. Suppose there are two or more points in $[-1,1]$ for which $x^3 - 3x + b = 0$ . Let $x_1, x_2$ be two such points with $x_1 < x_2$ . Since $x_1, x_2$ are both in $[-1,1]$ we have

$-1 \leq x_1 < x_2 \leq 1 \qquad \text{and} \qquad f(x_1) = f(x_2) = 0.$

Since $f(x) = x^3 - 3x + b$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$ (all polynomials are continuous and differentiable everywhere), we may apply Rolle’s theorem on the interval $[x_1, x_2]$ to conclude that there is some $c \in (x_1, x_2)$ such that

$f'(c) = 0 \quad \implies \quad 3c^2 - 3 = 0 \quad \implies \quad c = \pm 1.$

But this contradicts that $c \in (-1,1)$ (since $c \in (x_1, x_2)$ and $-1 \leq x_1 < x_2 \leq 1$ ). Hence, there can be at most one point $x \in [-1,1]$ such that $f(x) = 0. \qquad \blacksquare$

Prove there is exactly one negative solution to an equation

by RoRi

September 5, 2015

Show there is exactly one $b < 0$ such that $b^n = a$ for $a < 0$ and $n$ an odd, positive integer.

Proof. Let $f(x) = x^n$ , and let $c < -1$ with $c < a < 0$ . Then, $c^n < c$ (for odd $n$ ) so $c^n < c < a < 0$ . Since $f(0) = 0$ and $f(c) = c^n$ , by the Intermediate Value Theorem, we know $f(x)$ takes every value between $c^n$ and 0 for some $x \in [c,0]$ . Thus, we know there exists $b \in [c,0]$ such that $f(b) = a$ (since $c^n < a < 0$ ). This implies $b^n = a$ for some $b \in [c,0]$ .

We know this solution is unique since $f$ is strictly increasing on the whole real line for odd $n. \qquad \blacksquare$