Values of $m$ such that $|x^2-2x+m| + 2x + 1$ has $3$ extrema

Question

I was given the following question:

Find the values of $m$ for which the curve $y=|x^2-2x+m| + 2x + 1$ has $3$ extrema.

My teacher suggested that we should use the quadratic formula $(b^2-4ac)$ and check every case of it.

I am sure that there are three main cases of the inside of absolute value being $<0 ; =0 ; >0$.

The solution is $m<0$ but I do not know how to dig down into each case.

Answer 1

Note that your absolute value can be written as a piecewise function:

$$y = \begin{cases} x^2 - 2x + m + 2x + 1 = x^2 + m + 1 \\ -(x^2 - 2x + m) + 2x + 1 = 4x - x^2 -m + 1 \end{cases}$$

The domain of the piecewise function is determined by the value of $m$. I hope you understand why the piecewise function exists like such. If not, see the Absolute Value Function.

Consider, $x^2 - 2x + m > 0 \implies x^2 - 2x + m + 2x + 1 > 2x + 1, \forall x \in \mathbb{R} \iff m > 1$. Then, the absolute value would be redundant, and your function would be an ordinary +parabola; $x^2 + m + 1 \ge x^2 + 2$. Hence, there would not be $3$ extrema.

Therefore, our first condition for $3$ extrema must be that there are $2$ real solutions to the function, $-(x^2 - 2x + m) + 2x + 1$, since the $2$ of the extrema are these $2$ solutions.

Let these solutions be $x_1, x_2, x_1 \le x_2$.

$$x_1,x_2 = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\x_1 = \dfrac{-4 - \sqrt{4^2 - 4(-1)(1 - m)}}{2(-1)} = 2 - \sqrt{5 - m} \\x_2 = \dfrac{-4 + \sqrt{4^2 - 4(-1)(1 - m)}}{2(-1)} = 2 + \sqrt{5 - m}$$

We can then refine the domain to,

$$y = \begin{cases} x^2 - 2x + m + 2x + 1 = x^2 + m + 1 &x < x_1 \text{ or } x > x_2 \\ -(x^2 - 2x + m) + 2x + 1 = 4x - x^2 -m + 1 &x_1 < x < x_2 \end{cases}$$

The $3$rd extrema occurs when there exists a maxima between $x_1, x_2$. This is a maxima because the line $2x + 1$ is strictly increasing; $2x_1 + 1 < 2x_2 + 1 \text{ or } y(x_1) < y(x_2)$. Additionally, observe from above that $|x^2 - 2x + m| + 2x + 1 \ge 2x + 1$. The maxima can be calculated through the derivative:

$\big(-(x^2 - 2x + m) + 2x + 1\big)' = 0 \\4 - 2x = 0 \implies x = 2 \\\therefore 2x_2 + 1 = 2(2) + 1 = 5$

We must have $x_2 > 2, y(x_2) > 5$ (observing $2x + 1$ strictly increasing),

$$ \\\therefore 4x_2 + 1 - x_2^2 - m = 4(2) + 1 - 2^2 - m > 5 \\5 - m > 5 \\m < 0$$

Answer 2

It is easier to tackle this problem if we understand how the function behaves on the graph.

The curve is the sum of an absolute quadratic function and a linear function. Absolute of a quadratic Linear Sum of the two functions

It is apparent that when there are three extrema, they occur at the following points:

(1) The two points where the quadratic function hits the $x$-axis.

(2) The vertex of the quadratic function.

There are two conditions to fulfil.

(1) The quadratic function intersects the $x$-axis at two points. Otherwise, the absolute function cannot create the two kinks.

(2) The curve goes to opposite direction on the two sides of each kink. A counter-example is:

The slopes on both sides of the kink are upward so it is not considered an extrema.

$$\\$$ Solution:

To fulfil condition (1), we may use the discriminant. $$b^2 - 4ac > 0$$ $$(-2)^2 - 4m > 0$$ $$m < 1$$

To fulfil condition (2), we may check the derivatives on the two sides of each kink. The kinks occur at $$x^2 - 2x + m = 0$$ $$x = 1 \pm \sqrt{1-m}$$

Case I: $x < 1-\sqrt{1-m}$

$x^2-2x+m>0$

Hence, the curve can be simplified as $y = (x^2-2x+m)+(2x+1) = x^2+(m+1)$.

Case II: $1-\sqrt{1-m}<x<1+\sqrt{1-m}$

$x^2-2x+m<0$

Hence, the curve can be simplified as $y = -(x^2-2x+m)+(2x+1) = -x^2+4x+(1-m)$.

Case III: $x > 1+\sqrt{1-m}$

$x^2-2x+m>0$

Hence, the curve can be simplified as $y = x^2+(m+1)$. $$\\$$ The derivative of $y = x^2+(m+1)$ is given by $y' = 2x$.

The derivative of $y = -x^2+4x+(1-m)$ is given by $y' = -2x+4$. $$\\$$ That means, we want $2x$ and $(-2x+4)$ to have opposite signs at $x = 1 \pm \sqrt{1-m}$.

It requires that $$\begin{cases}2(1-\sqrt{1-m})<0 \\ -2(1-\sqrt{1-m})>0 \\ -2(1+\sqrt{1-m})<0 \\ 2(1+\sqrt{1-m})>0\end{cases}$$

The set of inequalities boil down to $$\begin{cases}\sqrt{1-m}>-1\\ \sqrt{1-m}>1\end{cases}\Rightarrow\sqrt{1-m}>1\Rightarrow m<0$$

As we need both conditions (1) & (2) to satisfy, i.e. $m < 1$ and $m < 0$, the answer to your question is $m<0$.

Answer 3

Here is a solution which only ever considers two cases.

We will use the fact that the derivative of $g(u)=|u|$ is $$ \frac{d}{du} g(u) = \left\{ \begin{array}{lr} 1 & \text{for } u>0\\ -1 & \text{for } u<0\\ \end{array} \right\} =: \text{sgn}(u) $$

Let $h(x) = x^2-2x+m$, and differentiate

$$ \begin{align} f'(x)&=\frac{d}{dx}\left(g(h(x))+2x+1\right) \\ &= h'(x)g'(h(x)) + 2\\ &= (2x-2)\text{sgn}(x^2-2x+m)+2 \\ &= \left\{ \begin{array}{lr} 2x & \text{for } x^2-2x+m>0 \cdots \text{(case I)}\\ 4-2x & \text{for } x^2-2x+m<0 \cdots \text{(case II)}\\ \end{array} \right. \end{align} $$

To get extrema, we want $f'(x)$ to change sign. Let's look at each case: Case I changes sign at $0$ and case II changes sign at $2$, so we make a table of the signs: $$ \begin{array}{lc|c|c} x^2-2x+m \in & \text{A}=(-\infty, 0) & \text{B}=(0,2) & \text{C}=(2,\infty) \\ f'(x) \text{ case I} & - & + & + \\ f'(x) \text{ case II} & + & + & - \\ \end{array} $$ Notice that if we only stay in case I, we will only get $1$ extremum. So we want to generate more extrema by choosing $m$ that allows $f'(x)$ to go between cases like I $\downarrow$ II $\uparrow$ I.

To do this, we try to solve $x^2-2x+m=0$. By the quadratic formula these are $1\pm \sqrt{1-m}$. So the roots (if any) are evenly spaced around $1$ by $\sqrt{1-m}$. Follow the table from left to right and label any extrema we find by E$1$ E$2,\ldots$, to see

$$ \begin{align} \sqrt{1-m} > 1 &\iff m < 0 \iff \text{ I} \downarrow \text{II in A (E1); } \ + \to - \text{ around } 2 \text{ (E2); II} \uparrow \text{I in C (E3)}\\ \sqrt{1-m} \le 1 &\iff m \ge 0 \iff \ - \to + \text{ around } 0 \text{ (E1); (and that is all)} \end{align} $$

So $f(x)$ has three extrema if and only if $m<0$.