Find the smallest value of positive constant $m$ such that curve will never lie below $x$-axis?

Question

Question:

Find the smallest value of positive constant $m$ that will make curve

$$y=mx-1+\dfrac{1}{x}$$

Greater than or equal to zero for all positive values of $x$?

My attempt:

By $AM\geq GM$

$mx+ \dfrac {1}{x} \geq 2.\sqrt{m}$

so,

for curve to attain positive or zero value

$2\sqrt m -1\geq 0\implies m\geq1/4$

giving minimum value $m=0.25$.

But I want to ask how to do this problem using only calculus without using inequalities because in my sheet, this problem was under the calculus section so I want to do it using calculus only. Thank you

Answer 1

A purely mechanical way would be to find the minimum value of $f(x)$ and find the $m$ that satisfy $f(x) \ge 0$, where $f(x) = mx -1 + \frac 1x$ and $x>0$.

$$\begin{align*} f'(x) &= m -\frac{1}{x^2}\\ f''(x) &= \frac{2}{x^3}\ge 0 \end{align*}$$

Minimum of $f(x)$ is at $f'(x) = 0$,

$$\begin{align*} m &= \frac{1}{x^2}\\ x &= \frac1{\sqrt{m}}\\ f\left(\frac1{\sqrt m}\right)&\ge 0\\ \sqrt m -1 + \sqrt m &\ge 0\\ \sqrt m &\ge \frac12\\ m &\ge \frac14 \end{align*}$$

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Though the question and the calculation above consider only $m > 0$, if $m\le 0$, $$f(2) = 2m - 1 + \frac12 = 2m - \frac{1}{2} \le -\frac 12 < 0$$

So no $m\le 0$ satisfies the condition.

Answer 2

Note: $m>0.$

$ mx -1+1/x >0.$

Since $x>0:$

$mx^2 -x +1>0.$

Completing the square:

$m(x^2-(1/m)x) +1 >0.$

$m[(x-1/(2m))^2 -1/(4m^2)] +1>0.$

$m(x-1/(2m))^2 -1/(4m) +1>0.$

The first term, a square, is greater or equal 0.

Hence:

$-1/(4m)+1>0$, or

$4m>1,$ or $ m>1/4$.

Answer 3

Imagine you have a string in the Cartesian plane that is mostly above the x-axis, but some of it is below. Now imagine this string moves in such a way that all of it is above the x-axis. Now think about what has to happen at the moment it goes from being partly below to being completely above. At that moment, it has to be tangent to the x-axis. That is, it has to have a point with y-value of zero (because it's right on the edge between being below the x-axis and not), and the slope has to be zero there (if, for instance, the slope were positive, then that would mean that there's a little bit to the left that's below, so you could still move it a little more while keeping some of it below the x-axis).

So you're looking for a point where both y and y' are zero. First, set y' to zero.

y'= m-1/x² = 0

x = $\frac{1}{\sqrt{m}}$

Next, set y to zero

y = mx -1 + 1/x = $\sqrt{m} -1 + \sqrt{m} $ = 0

So

$2 \sqrt{m} = 1$

$\sqrt{m} = \frac{1}{2}$

$m = \frac{1}{4}$

Another way to think of it:

Imagine that the string intersects the x-axis at r₁ and r₂. As less and less of the string is below the x-axis, r₁ and r₂ should get closer and closer together. You're looking for the moment where they are the same.

You have

y = mx -1 + 1/x

y = (mx² - x +1)/x

Setting this to zero, you have

(mx² - x +1)/x = 0

mx² - x +1 = 0

You now want the discriminant in the quadratic formula to be zero so that the two roots will be the same.

b²-4ac = 0

1-4m = 0

m = 1/4