A function with three local extrema¶

Math I: Calculus · Mock exam, Question 3(b) · 14 points · Lectures 6–7

Problem¶

Let

g(x)=\mathrm{e}^{-x}+a\sqrt{x},\quad x\in[0,\infty),

with $a\in\mathbb{R}\text{,}$ and suppose that $g$ has exactly three local extrema, at $0=x_0<x_1<x_2\text{.}$

(b1) Find $g'(x)$ for $x>0\text{.}$ Determine for which values of $a$ a finite right derivative exists at $0\text{.}$ [4 points]
(b2) Determine all possible values of $a$ under the stated assumption of three local extrema. [6 points]
(b3) Prove that $x_2+g(x_1)>3/2\text{.}$ [4 points]

What you need¶

One-sided derivative: the right difference quotient at an endpoint.
Exponential function: $(\mathrm{e}^{-h}-1)/h\to-1$ as $h\to0\text{.}$
Stationary point and local extremum, including extrema at an endpoint of the domain.
Monotone function: the sign of a derivative gives where a function increases or decreases.

Plan¶

Separate the parameter from $x\text{.}$ For $x>0\text{,}$

g'(x)=\frac{a-2\sqrt{x}\,\mathrm{e}^{-x}}{2\sqrt{x}},

so the sign of $g'$ is the sign of $a-h(x)$ with $h(x)=2\sqrt{x}\,\mathrm{e}^{-x}\text{.}$ Counting stationary points becomes counting how often the horizontal line $y=a$ meets the graph of $h\text{.}$ For (b3), use the stationary-point equation to remove $a$ from $g(x_1)\text{.}$

Step-by-step solution¶

Try each step yourself before you open it.

(b1) Step 1 · The derivative for x > 0 (1 point)

g'(x)=-\mathrm{e}^{-x}+\frac{a}{2\sqrt{x}}.

(b1) Step 2 · The right difference quotient at 0 (1 point)

Since $g(0)=1\text{,}$ for $h>0$

\frac{g(h)-g(0)}{h}=\frac{\mathrm{e}^{-h}-1}{h}+\frac{a}{\sqrt{h}}.

The formula from Step 1 is not valid at $x=0\text{;}$ at the endpoint you must use the definition.

(b1) Step 3 · Analyse both terms (2 points)

As $h\to0^+\text{,}$ the first term tends to $-1\text{,}$ the derivative of $\mathrm{e}^{-x}$ at $0\text{.}$ The second term $a/\sqrt{h}$ has a finite limit only when $a=0\text{.}$ So a finite right derivative exists exactly for $a=0\text{,}$ and then it equals $-1\text{.}$

Concept: One-sided derivative

(b2) Step 1 · Separate the parameter (1 point)

For $x>0\text{,}$ $g'(x)=0$ is equivalent to

a=h(x):=2\sqrt{x}\,\mathrm{e}^{-x}.

(b2) Step 2 · Study h (3 points)

Differentiate:

h'(x)=\frac{\mathrm{e}^{-x}}{\sqrt{x}}(1-2x).

So $h$ increases on $(0,1/2)$ and decreases on $(1/2,\infty)\text{.}$ Its maximum is

h\left(\tfrac{1}{2}\right)=\sqrt{2}\,\mathrm{e}^{-1/2}=\sqrt{2/\mathrm{e}}.

Also $h(x)\to0$ as $x\to0^+$ and as $x\to\infty\text{,}$ and $h>0\text{.}$

(b2) Step 3 · Count solutions (1 point)

The equation $h(x)=a$ has exactly two positive solutions precisely when

0<a<\sqrt{2/\mathrm{e}}.

For $a\leq0$ there is none, for $a=\sqrt{2/\mathrm{e}}$ only $x=1/2\text{,}$ and for larger $a$ none.

(b2) Step 4 · Check that all three are extrema (1 point)

For $0<a<\sqrt{2/\mathrm{e}}$ the solutions satisfy $0<x_1<1/2<x_2\text{.}$ Because $g'(x)$ has the sign of $a-h(x)\text{,}$ $g'$ is positive on $(0,x_1)\text{,}$ negative on $(x_1,x_2)$ and positive on $(x_2,\infty)\text{.}$ So $x_1$ is a local maximum and $x_2$ a local minimum. At the endpoint, $g$ increases immediately to the right of $0\text{,}$ so $0$ is a local minimum relative to the domain.

The other values of $a$ fail: for $a\leq0\text{,}$ $g'<0$ on $(0,\infty)\text{;}$ for $a\geq\sqrt{2/\mathrm{e}}\text{,}$ $g'\geq0$ with at most one zero. In both cases $0$ is the only local extremum. Hence the possible values are exactly $0<a<\sqrt{2/\mathrm{e}}\text{.}$

(b3) Step 1 · Use x₂ > 1/2 (1 point)

From (b2), $x_2>1/2\text{.}$

(b3) Step 2 · Remove a from g(x₁) (1 point)

At $x_1$ the stationary-point relation gives $a=2\sqrt{x_1}\,\mathrm{e}^{-x_1}\text{,}$ so $a\sqrt{x_1}=2x_1\mathrm{e}^{-x_1}$ and

g(x_1)=(1+2x_1)\mathrm{e}^{-x_1}.

(b3) Step 3 · A monotone auxiliary function (1 point)

Let $q(x)=(1+2x)\mathrm{e}^{-x}\text{.}$ Then

q'(x)=(1-2x)\mathrm{e}^{-x}>0\quad\text{on }(0,\tfrac{1}{2}),

so $q$ is strictly increasing on $[0,1/2]\text{.}$ Since $0<x_1<1/2\text{,}$ $g(x_1)=q(x_1)>q(0)=1\text{.}$

(b3) Step 4 · Combine (1 point)

x_2+g(x_1)>\frac{1}{2}+1=\frac{3}{2}.

Where marks are lost¶

Using the formula for $g'$ at $x=0\text{.}$ The endpoint needs the difference quotient.
Finding $0<a<\sqrt{2/\mathrm{e}}$ without checking the sign pattern. Two stationary points are not automatically two extrema, and the endpoint extremum must be argued.
Leaving $a$ in $g(x_1)\text{.}$ The stationary-point equation is what turns (b3) into a one-variable inequality.

Try next¶

Symbols: element of, infinity, real numbers, square root