Inequality

[Randyyy]

Hey,
Question: What values for k satisfies the given inequality [MATH]ln(2x) \leq kx \leq e^{x/2} [/MATH] for x > 0
I started by noticing that ln(2x) is the inverse to [MATH]e^{x/2}[/MATH], so the line kx has to be between the two curves, which is also obvious perhaps from the given inequality. My first thought was to try and calculate kx as a tangent to ln(2x) and [MATH]e^{x/2}[/MATH] but neither have a maxima so I can´t target with a tangent and make kx be between those x coordinates. Certainly I can guess that for example k = 1 satisfies is but surely there is an algebraic way to go about this?

Thanks in advance!

 

[Dr.Peterson]

Hey,
Question: What values for k satisfies the given inequality [MATH]ln(2x) \leq kx \leq e^{x/2} [/MATH] for x > 0
I started by noticing that ln(2x) is the inverse to [MATH]e^{x/2}[/MATH], so the line kx has to be between the two curves, which is also obvious perhaps from the given inequality. My first thought was to try and calculate kx as a tangent to ln(2x) and [MATH]e^{x/2}[/MATH] but neither have a maxima so I can´t target with a tangent and make kx be between those x coordinates. Certainly I can guess that for example k = 1 satisfies is but surely there is an algebraic way to go about this?

Thanks in advance!

You have a valid idea in saying "try and calculate kx as a tangent to ln(2x) and [MATH]e^{x/2}[/MATH]", but it has nothing to do with maxima. You want to find the line through the origin that is tangent to each curve; so set the derivative at x (the slope of the curve) equal to y/x (the slope of the line through the origin) and solve for x.

 

[Steven G]

Isn't it obvious that y=x must be the answer if there is an answer of the form y=kx?
I just realized that you said that k=1 is an obvious answer. Good job.

 

[Randyyy]

You have a valid idea in saying "try and calculate kx as a tangent to ln(2x) and [MATH]e^{x/2}[/MATH]", but it has nothing to do with maxima. You want to find the line through the origin that is tangent to each curve; so set the derivative at x (the slope of the curve) equal to y/x (the slope of the line through the origin) and solve for x.

I let[MATH]u(x) = ln(2x) [/MATH] and [MATH]y(x)=e^{x/2}[/MATH]and solved the following equations, [MATH]\frac{u(x)-0}{x-0}=u'(x)[/MATH] and [MATH]\frac{y(x)-0}{x-0}=y'(x)[/MATH], these gave me the solutions:
[MATH]x_1 = 2, x_2 = \frac{e}{2}[/MATH], [MATH]u'(x_1)=k_1, y'(x_2)=k_2 \implies k_1=\frac{e}{2}, k_2=\frac{2}{e}[/MATH]I might have written them in the wrong order because I wasn´t paying too much attention.

I used https://www.desmos.com/calculator/b5uwdwxnie to confirm my answer and so my final answer is that [MATH]\frac{2}{e}\leq k \leq \frac{e}{2}[/MATH]
Thanks for taking your time to assist me!

 

[Dr.Peterson]

I let[MATH]u(x) = ln(2x) [/MATH] and [MATH]y(x)=e^{x/2}[/MATH]and solved the following equations, [MATH]\frac{u(x)-0}{x-0}=u'(x)[/MATH] and [MATH]\frac{y(x)-0}{x-0}=y'(x)[/MATH], these gave me the solutions:
[MATH]x_1 = 2, x_2 = \frac{e}{2}[/MATH], [MATH]u'(x_1)=k_1, y'(x_2)=k_2 \implies k_1=\frac{e}{2}, k_2=\frac{2}{e}[/MATH]I might have written them in the wrong order because I wasn´t paying too much attention.

I used https://www.desmos.com/calculator/b5uwdwxnie to confirm my answer and so my final answer is that [MATH]\frac{2}{e}\leq k \leq \frac{e}{2}[/MATH]
Thanks for taking your time to assist me!

Good work. You didn't do it exactly the way I suggested, which means you're thinking for yourself, which is good.

But in case anyone else reads this, we should fix a typo: You meant [MATH]x_1 = \frac{e}{2}, x_2 = 2[/MATH], right? Please do pay attention when you write explanations for others to read!

 

[Dr.Peterson]

Isn't it obvious that y=x must be the answer if there is an answer of the form y=kx?
I just realized that you said that k=1 is an obvious answer. Good job.

Be careful to distinguish between "the answer" and "an answer". y=x is the latter, not the former; the former is what Randyy has given.