Showing the inequality holds for an interval (?)

[life is maths]

Showing the inequality holds for an interval (?)

Homework Statement

Hi, my homework question is:

Show that the inequality
$\sqrt{2+x}$<2+$\frac{x}{4}$ holds $\forall$x$\in$[-2,0]

Homework Equations

The Attempt at a Solution

I tried using IVT or bisection method, but they are just for existence of a root. How can I show it holds for all x in the interval [-2,0]? Would taking the derivative of the function
$\sqrt{2+x}$-2-$\frac{x}{4}$ lead me anywhere? Like finding maximum or minimum points? Thanks a lot for any help.

 

[Dick]

Find the roots of sqrt(2+x)-(2-x/4)=0. The use the roots to sketch a graph. Where your graph is negative is where sqrt(2+x)<(2-x/4), right?

 

[life is maths]

Find the roots of sqrt(2+x)-(2-x/4)=0. The use the roots to sketch a graph. Where your graph is negative is where sqrt(2+x)<(2-x/4), right?

Thanks, Dick, but when I try finding roots of $\sqrt{2+x}$-(2+$\frac{x}{4}$)= 0, what I get is:

2+x = [$\frac{x}{4}$]²+4+x
x²/16+2=0
but it is not possible. How should I proceed from here? Is there a mistake?

 

[Dick]

Thanks, Dick, but when I try finding roots of $\sqrt{2+x}$-(2+$\frac{x}{4}$)= 0, what I get is:

2+x = [$\frac{x}{4}$]²+4+x
x²/16+2=0
but it is not possible. How should I proceed from here? Is there a mistake?

No, no mistake. If there are no roots or singularities then that difference must always be positive or always be negative since it doesn't cross zero. Which is it?

 

[life is maths]

x²/16+2 must always be positive for all intervals of x. If we put the inequality sign instead of 0, the statement is true. Should I express it like: 'Since x²/16+2>0 is true for all x, then it also holds for x$\in[-2,0]$' ? Is this a formal statement?

 

[Mark44]

That's formal enough for me.

 

[life is maths]

Thanks a lot, I had thought I should solve for a specific case, but this is also true :)

 

[Mark44]

If you want to get fancy, you can say this: Since the inequality is true for all real numbers, it's true a fortiori, for x $\in$ [-2, 0].

Loosely translated, the italicized phrase means "even more strongly."

 

[Dick]

Thanks a lot, I had thought I should solve for a specific case, but this is also true :)

I think you should. So far you've only shown the curves sqrt(2+x) and 2+x/4 don't cross, since they are never equal. So one is always greater than the other one. To say WHICH one is greater, I would test them at, say, x=(-2).

 

[life is maths]

I think you should. So far you've only shown the curves sqrt(2+x) and 2+x/4 don't cross, since they are never equal. So one is always greater than the other one. To say WHICH one is greater, I would test them at, say, x=(-2).

Thanks, Dick. If I take x= -2, then I get $\sqrt{-2+2}$< 2+ $\frac{(-2)}{4}$, which is 0 < $\frac{3}{2}$. It also holds for 0, then $\sqrt{2+x}$ < 2+$\frac{x}{4}$ in the interval [-2,0]. Now how should I proceed? How can I show this holds for every number in this interval?
I'm a bit confused

 

[Dick]

Thanks, Dick. If I take x= -2, then I get $\sqrt{-2+2}$< 2+ $\frac{(-2)}{4}$, which is 0 < $\frac{3}{2}$. It also holds for 0, then $\sqrt{2+x}$ < 2+$\frac{x}{4}$ in the interval [-2,0]. Now how should I proceed? How can I show this holds for every number in this interval?
I'm a bit confused

Because you showed the graphs don't cross because there are no values where they are equal. So sqrt(x+1) is ALWAYS greater than 2+x/4 for any x>=(-2). It's true on [-2.infinity). So it must also be true on [-2,0].

 

[life is maths]

Wow, drawing their graphs shows everything clearly. sqrt(2+x) is always smaller than 2+x/4. Hence, this is true for every interval that they are defined (x>=-2).

Thanks for your time and effort, Dick and Mark :) You were a great help.