What are some good ways to find maximum for $d=\frac {\left| m+1\right| }{\sqrt {m^{2}+1}}$?

Question

I have been having some difficulties with this question.

How to find the maximum without the help of a calculator or graphing device?

Answer 1

Hint: $$d^2=\dfrac {(m+1)^2} {m^{2}+1}=1+2\frac{m}{m^2+1}.$$ Note that for $m>0$, $$\frac{m}{m^2+1}=\frac{1}{m+\frac1m}\leq \frac{1}{2}$$ since $m+\frac1m\geq2$ for $m>0$.

Answer 2

For each $m$, Consider the line $\ell_m$ given by $y+mx=0$. Observe that $\ell_m$ passes through the origin. Also consider the point $(1,1)$. Then the distance of the line $\ell_m$ from the point $(1,1)$ is given by $$d=\frac{|m+1|}{\sqrt{m^2+1}}.$$ So we want that line whose distance from the point $(1,1)$ will be maximum. Since $\ell_m$ passes through the origin, therefore the maximum distance will occur when the line joining the origin and the point $(1,1)$ is perpendicular to line $\ell_m$. This will occur when $m=1$. So the maximum value of $d$ is $\color{red}{\sqrt{2}}$.

Note:(for more clarity)

Let $P$ be the point $(1,1)$, $O$ be the origin. Then consider the line $\ell_m$ with $\color{blue}{m \neq 1}$, then the distance between $P$ and this $\ell_m$ is obtained by the perpendicular that can be dropped from $P$ onto $\ell_m$, call the foot of this perpendicular to be $A$. Observe that $\triangle PAO$ is right angled and $PO$ is the hypotenuse, so $PO>PA$. Thus the maximum occurs with the line mentioned above.

Answer 3

WLOG $m=\tan t$ where $-\dfrac\pi2<t<\dfrac\pi2\implies\sec t=+\sqrt{1+m^2}$

$$d=\dfrac{\tan t+1}{\sec t}=\sin t+\cos t=\sqrt2\sin\left(t+\dfrac\pi4\right)\le\sqrt2$$

Answer 4

As $d\ge0$

$$d=\dfrac{|m+1|}{\sqrt{m^2+1}}\iff d^2(m^2+1)=(m+1)^2$$

$$\iff(d^2-1)m^2-2m+(d^2-1)=0$$

As $m$ is real, the discriminant must be $\ge0$

$$2^2\ge4(d^2-1)^2\iff0\ge(d^2-1)^2-1^2=(d^2-2)d^2\iff0\le d^2\le2$$

Answer 5

By inequality beetwen quadratic and arithmetic mean we have $$\sqrt{m^2+1\over 2}\geq {|m|+1\over 2}$$ we have $${|m|+1\over \sqrt{m^2+1}} \leq \sqrt{2}$$

But $|m+1|\leq |m|+1$ so $d\leq \sqrt{2}$ and that is achieved if $m= 1$