Solution of inequality of composite function involving inverse

[songoku]

Homework Statement

Let $f(x)=x^4+0.2x^3-5.8x^2-x+4, 0 \leq x \leq 1.67$ and $g(x)=2 \sin(x-1) -3 , -\frac{\pi}{2}+1 \leq x \leq \frac{\pi}{2}+1$

(i) Find $f^{-1} (x)$
(ii) Solve $(f^{-1} o~ g)(x) < 1$

Relevant Equations

Inverse and composite function

I can solve (i), I got x = -1.6

For (ii), I did like this:
$$(f^{-1} o ~g)(x)<1$$
$$g(x)<f(1)$$

But it is wrong, the correct one should be $g(x) > f(1)$. Why?

Thanks

 

[fresh_42]

Firstly: What is $f^{-1}$? Is it the solution of $f^{-1}\circ f =\operatorname{id}$ or is it the set $f^{-1}=\{y\in \mathbb{R}\, : \,y=f(x)\text{ for some }x\}?$

Secondly: If it is the former, the inverse function, then how can it be a single value? Shouldn't it be another function?

You can divide $f(x)$ by $(x+1)$ and $(x-0.8)$ for a better handling.

 

[songoku]

Firstly: What is $f^{-1}$? Is it the solution of $f^{-1}\circ f =\operatorname{id}$ or is it the set $f^{-1}=\{y\in \mathbb{R}\, : \,y=f(x)\text{ for some }x\}?$

Secondly: If it is the former, the inverse function, then how can it be a single value? Shouldn't it be another function?

I am so sorry I mistyped the question. Question (i) should be: solve $f^{-1} (x) = 1$ so I did it like this:
$$f^{-1} (x) =1$$
$$x= f(1)$$
$$x = -1.6$$

$f^{-1} (x)$ is inverse function of f(x)

 

[Office_Shredder]

When you apply f to both sides of an inequality, that only preserves the inequality if f is an increasing function. If f is decreasing, if flips the inequality (e.g. multiply both sides by -1). If f is neither, then you can't really say much about the inequality after applying f to both sides.

 

[fresh_42]

It is strictly increasing and one-to-one on the given interval.

If you are looking for an $x$ such that $f(x)=1$, then we have to solve
$$
g(x):=x^4+0.2x^3-5.8x^2-x+3=0
$$
This equation doesn't have 'nice' solutions. Are you expected to find them by an algorithm, or graphically? $x=-1.6$ is wrong and furthermore not in the given interval.

If you want to calculate $f(1)$ then the decomposition of $f(x)$ into irreducible factors is useful. I already told you two zeros: $-1$ and $0.8$ so you can divide $f(x)$ by $(x+1)$ and $(x-0.8)$ which allows you a fast calculation of any value of $f(x)$.

 

[Office_Shredder]

Fresh, f(1) is trivial to solve, I don't see why you need to factor it. And I think it is -1.6.

Also, f is clearly decreasing at 0 at least, since the linear term dominates there and is negative sloped.

 

[fresh_42]

Homework Statement:: Let $f(x)=x^4+0.2x^3-5.8x^2-x+4, 0 \leq x \leq 1.67$

Fresh, f(1) is trivial to solve, I don't see why you need to factor it. And I think it is -1.6.
Also, f is clearly decreasing at 0 at least, since the linear term dominates there and is negative sloped.

I used $x^4+0.2*x^3-5.8*x^2-x+4=1$ in WolframAlpha and got $0.67.$ Yes, factorization may not be necessary, but I use every chance I get to make students practice long division. And knowing the zeroes helps to visualize the function.

 

[Office_Shredder]

Fresh, you've made things too complicated. f(x) is the polynomial we have already been given. We want to solve for f(1). It looks like you're trying to find $f^{-1}(1)$

 

[fresh_42]

Fresh, you've made things too complicated. f(x) is the polynomial we have already been given. We want to solve for f(1). It looks like you're trying to find $f^{-1}(1)$

Yes, that's what I thought and why I asked for the meaning of $f^{-1}$. It exists on the given interval, and to actually invert it there is more complicated than to calculate $f(1)$.

 

[Office_Shredder]

Maybe you missed post #3? They didn't need to compute $f^{-1}$ for the first part.

 

[songoku]

Firstly: What is $f^{-1}$? Is it the solution of $f^{-1}\circ f =\operatorname{id}$ or is it the set $f^{-1}=\{y\in \mathbb{R}\, : \,y=f(x)\text{ for some }x\}?$

I encountered these two terms a few times but I still don't understand the difference. What is the difference between $f^{-1}$ as inverse function and set $f^{-1}=\{y\in \mathbb{R}\, : \,y=f(x)\text{ for some }x\}?$ ?

When you apply f to both sides of an inequality, that only preserves the inequality if f is an increasing function. If f is decreasing, if flips the inequality (e.g. multiply both sides by -1). If f is neither, then you can't really say much about the inequality after applying f to both sides.

What is the logic behind this? Why when applying f to both sides we need to consider increasing and decreasing function?

Thanks

 

[Office_Shredder]

What is the logic behind this? Why when applying f to both sides we need to consider increasing and decreasing function?

Thanks

An increasing function by definition is one for which if $x< y$, then $f(x)< (y)$, or to put it in words, as x increases, $f(x)$ increases. A decreasing function goes the other way, if $x<y$, then $f(x) > f(y)$.

It helps to think of the functions $f(x)=2x$ and $f(x)=-x$. You should already be familiar that applying the first one to both sides of an inequality preserves the inequality, and applying the second one flips it.

 

[fresh_42]

I encountered these two terms a few times but I still don't understand the difference. What is the difference between $f^{-1}$ as inverse function and set $f^{-1}=\{y\in \mathbb{R}\, : \,y=f(x)\text{ for some }x\}?$ ?

Consider $f(x)=|x|$. It has no global inverse function, but you can define its pre-images: $f^{-1}(1)=\{\pm 1\}$ or $f^{-1}(\mathbb{R}_{>0})=\mathbb{R},f^{-1}(\mathbb{R}_{<0})=\emptyset .$

If we restrict ourselves to the nonnegative part of the domain, $f:\mathbb{R}{>0} \longrightarrow \mathbb{R}\, , \,f(x)=|x|=x,$ then it has an inverse: $f^{-1}=f$ which is itself in this case.

So the basic difference is, that one is a function, and the other one is a set.

 

[songoku]

Consider $f(x)=|x|$. It has no global inverse function, but you can define its pre-images: $f^{-1}(1)=\{\pm 1\}$ or $f^{-1}(\mathbb{R}_{>0})=\mathbb{R},f^{-1}(\mathbb{R}_{<0})=\emptyset .$

If we restrict ourselves to the nonnegative part of the domain, $f:\mathbb{R}{>0} \longrightarrow \mathbb{R}\, , \,f(x)=|x|=x,$ then it has an inverse: $f^{-1}=f$ which is itself in this case.

So the basic difference is, that one is a function, and the other one is a set.

I think I understand it a bit more now. So the notation $f^{-1}$ does not always mean inverse function

Thank you very much for the help and explanation fresh_42 and Office_Shredder