Proving Inequality: Solving for x in x+3^x<4

[SebastianBS]

Find all numbers x for wich:
$$x+3^x<4$$
Relevant equations
(PI) (Associative law for addition)
(P2) (Existence of an additive identity)
(P3) (Existence of additive inverses)
(P4) (Commutative law for addition)
(P5) (Associative law for multiplication)
(P6) (Existence of a multiplicative identity)
(P7) (Existence of multiplicative inverses)
(P8) (Commutative law for multiplication)
(P9) (Distributive law)
(P10) (Trichotomy law)
(P11) (Closure under addition)
(P12) (Closure under multiplication)
THEOREM l For all numbers a and b, we have
$$\mid{a+b}\mid\leq\mid{a}\mid+\mid{b}\mid$$
I tried everything I'd already done with all the other problems but I just can't figure it out. Note that what I'm afteris the prove, I know what the result is. if you want look at it:

Spoiler

x<1

Edited: sorry mark, fixed it

 

[Mark44]

Find all numbers x for wich:
$$x+3^x$$

For which numbers x does x + 3^x do what?

This isn't an equation, and it isn't an inequality (inequation is not a word), so you can't solve it for x.

What is the complete inequality?

Relevant equations
(PI) (Associative law for addition)
(P2) (Existence of an additive identity)
(P3) (Existence of additive inverses)
(P4) (Commutative law for addition)
(P5) (Associative law for multiplication)
(P6) (Existence of a multiplicative identity)
(P7) (Existence of multiplicative inverses)
(P8) (Commutative law for multiplication)
(P9) (Distributive law)
(P10) (Trichotomy law)
(P11) (Closure under addition)
(P12) (Closure under multiplication)
THEOREM l For all numbers a and b, we have
$$\mid{a+b}\mid\leq\mid{a}\mid+\mid{b}\mid$$
I tried everything I'd already done with all the other problems but I just can't figure it out. Note that what I'm afteris the prove, I know what the result is. if you want look at it:

Spoiler

x<3

 

[epenguin]

= is when it probably passes between < and > isn't it?

Seems to me you've done the most difficult part and the rest is pretty qualitative, does LHS increase, decrease...?

 

[SebastianBS]

= is when it probably passes between < and > isn't it?

Seems to me you've done the most difficult part and the rest is pretty qualitative, does LHS increase, decrease...?

What's in OP is all the given information

 

[Mark44]

Solving x + 3^x = 4 for the exact solution is not something that is taught in precalculus courses, or even most calculus courses. About the best you can do is to get an approximate solution to x + 3^x = 4 (either graphically or by some estimation technique), and then use that to determine the interval for which x + 3^x < 4.

 

[Curious3141]

The best you can do here with elementary techniques is to sketch the curve y = x + 3^x, prove it's monotone increasing throughout, then hazard an intelligent guess that when y = 4, x = 1, which is trivially proven by substitution.

 

[epenguin]

What's in OP is all the given information

Sorry, maybe I misread your spoiler in haste or you changed it, but for x = 1

x + 3^x = 4 .

That is what I call the hard bit.

I think we have severally indicated to you the rest.

 

[SebastianBS]

This is from the first chapter of calculus by spivak, anyone who have worked with this author please explain

 

[Curious3141]

This is from the first chapter of calculus by spivak, anyone who have worked with this author please explain

Yes, the problem is in Chapter 1 (Basic Properties of Numbers), No. 4 (xii).

Everything else in No. 4 is exactly solvable, except this one. There's nothing on this in the "Answers to selected problems" section either. I'm guessing this problem was included only because it has an "obvious" solution, which can be easily proven to be unique with elementary techniques like curve-sketching, which even a beginning student of Calculus is expected to be already familiar with.