Composition of two piecewise functions f and g

[nicoc]

Hello everyone. This is my first post so if something about what i said is wrong or doesnt follow the rules just tell me.

Im having difficulties with (g o f)(x) of the functions below:

. . . . .[math]f(x)\, =\, \begin{cases}3x^2\, +\, 1&\, x\, \leq\, 0\\x\, -\, 8&\,0\, <\, x\, <\, 2\\5&\, x\, \geq\, 2\end{cases}[/math]
. . . . .[math]g(x)\, =\, \begin{cases}x\, -\, 1&\, x\, \leq\, 0\\ -1&\,0\, <\, x\, <\, 2\\\frac{1}{x}&\, x\, \geq\, 2\end{cases}[/math]
Im confused about 3x^2 + 1 when x<=0 because it doesnt have a real root so i dont know what to do with that part of the function. I appreciate if someone helps me to find the solution.

Thanks in advance.

 

[Otis]

… with (g o f)(x) … I'm confused about 3x^2 + 1 when x<=0 because it doesnt have a real root …

Hello. We don't need to consider roots, when composing (g◦f)(x) in this exercise. You're right, when you say f has no root, but that simply means the number zero will not be going into function g, when x is zero or less. In fact, for x≤0, the restricted range (output) of 3x^2+1 is [1,∞). Hence, for x≤0, the number 1 is the smallest output of f going into g.

The composition (g◦f)(x) is what we get, when we use the output of function f as the input to function g.

Instead of thinking in terms of actual numbers, try working symbolically with the given piecewise expressions for functions f and g. Please show what you're able to try (whether you feel good about it or not), and we can go from there. Cheers

PS: Looking at the piecewise definitions, we see that g(x) is 1/x, for x≥2. That's the only interval of x where we would want to be concerned about the output of f. That is, we would not want an output of zero going into function g because that would lead to 1/0. But note! Function f never outputs zero, for any input. So, there's no issue with division by zero in this composition.

EDIT: Math typo (1,∞) corrected as [1,∞)

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[pka]

Hello everyone. This is my first post so if something about what i said is wrong or doesnt follow the rules just tell me.

Im having difficulties with (g o f)(x) of the functions below:

. . . . .[math]f(x)\, =\, \begin{cases}3x^2\, +\, 1&\, x\, \leq\, 0\\x\, -\, 8&\,0\, <\, x\, <\, 2\\5&\, x\, \geq\, 2\end{cases}[/math]
. . . . .[math]g(x)\, =\, \begin{cases}x\, -\, 1&\, x\, \leq\, 0\\ -1&\,0\, <\, x\, <\, 2\\\frac{1}{x}&\, x\, \geq\, 2\end{cases}[/math]
Im confused about 3x^2 + 1 when x<=0 because it doesnt have a real root so i dont know what to do with that part of the function. I appreciate if someone helps me to find the solution.

I will give you a series of examples:
\(\displaystyle g\circ f(-2)=g(13)=\frac{1}{13}\\g\circ f(-1)=g(4)=\frac{1}{4}\\g\circ f(0)=g(1)=-1\\g\circ f(1)=g(-7)=-8\\g \circ f(2)=g(5)=\frac{1}{5}\)
Now you post what you can do now.

 

[nicoc]

Thanks for the reply.

What I did was to get the range of every part of the function f.
3x^2+1 has range (1,+inf) when x<=0
x-8 has range(-8,-6) when 0<x<2
5 has range [5] when x>=2

When 0<x<2 the output of function f is <=0. So (g o f)(x)=x-9 when 0<x<2

The part that take me long to understand was, what to do with function 3x^2+1 and 5, because the output of both are greater or equal to 2 so that is a conflict with 1/x in function g because I can either put 1/(3x^2+1) or 1/5 in (g o f)(x).

Because of that I thought that the range of 3x^2+1 is [1,2) when -1/sqrt(3)<x<=0 UNION [2,+inf) when x<=-1/sqrt(3).
With that in mind:

(g o f)(x) =
1/(3x^2+1) when x<=-1/sqrt(3)
-1 when -1/sqrt(3)<x<=0
x-9 when 0<x<2
1/5 when x>=2

Is this right?

Thats all. Hope you can understand what I did and my english too. Next time I will try to use latex or upload an image.

 

[pka]

Thanks for the reply.

What I did was to get the range of every part of the function f.
3x^2+1 has range (1,+inf) when x<=0
x-8 has range(-8,-6) when 0<x<2
5 has range [5] when x>=2

When 0<x<2 the output of function f is <=0. So (g o f)(x)=x-9 when 0<x<2

The part that take me long to understand was, what to do with function 3x^2+1 and 5, because the output of both are greater or equal to 2 so that is a conflict with 1/x in function g because I can either put 1/(3x^2+1) or 1/5 in (g o f)(x).

Because of that I thought that the range of 3x^2+1 is [1,2) when -1/sqrt(3)<x<=0 UNION [2,+inf) when x<=-1/sqrt(3).
With that in mind:

(g o f)(x) =
1/(3x^2+1) when x<=-1/sqrt(3)
-1 when -1/sqrt(3)<x<=0
x-9 when 0<x<2
1/5 when x>=2

Is this right?

Thats all. Hope you can understand what I did and my english too. Next time I will try to use latex or upload an image.

Here is my take on a beast of a question:
\(\displaystyle \begin{cases}f(x)\le 0 &: 0<x<2 \\ 0<f(x)<2 &: -\frac{\sqrt{3}}{3}<x\le 0\\2\le f(x) &: x\le-\frac{\sqrt{3}}{3}\text{ or }2\le x \end{cases}\)
Actually I will welcome any & all corrections.

So what is \(\displaystyle g\circ f(x)=~?\)

 

[Otis]

… the range of every part of the function f …
3x^2+1 has range [1,+inf) when x<=0
x-8 has range(-8,-6) when 0<x<2
5 has range [5] when x>=2

Hello, again. That all looks good, but note the change shown in red above. I'd made the same typo, in my first reply (now fixed).

When 0<x<2 the output of function f is <=0 …

No, function f never outputs zero, in this exercise. Look at your ranges for f, again. None of those include 0.

When x is between 0 and 2, function f outputs negative values only.

… So (g o f)(x) = x-9 when 0<x<2

That's correct! Function g subtracts 1 from its input, whenever that input is negative. When 0<x<2, function f outputs the negative value (x-8), and that negative value then goes into function g, and so g subtracts 1 from it:

g(x-8) = (x-8) - 1 = x - 9

… what to do with function 3x^2+1 and 5, because the output of both are greater or equal to 2 …

That's not quite correct. For some values of x≤0, the output 3x^2+1 is ≥ 2. But, for the other values of x, the output 3x^2+1 is less than 2. Again, look at your range for f, when x≤0. Some of those values are <2 and the rest are ≥2. Therefore, you need to determine which negative values of x cause f(x) to be less than 2 and which cause f(x) to be 2 or more.

… Because of that I thought that the range of 3x^2+1 is [1,2) when -1/sqrt(3)<x<=0 … [2,+inf) when x<=-1/sqrt(3).

Oh, you already did what I just suggested. (I should have studied your entire post, first!)

With that in mind:
(g o f)(x) =
1/(3x^2+1), when x<=-1/sqrt(3)
-1, when -1/sqrt(3)<x<=0
x-9, when 0<x<2
1/5, when x>=2

I think you've got it. Good job!!

Depending on your instructor, you may want to consider writing -1/√3 as (-√3)/3.

This exercise has been a real brain-twister.

 

[Otis]

… Hope you can understand what I did and my english too. Next time I will try to use latex or upload an image.

Your English seems excellent, and I had no difficulty understanding your math statements. (We both made some typos, I think.)

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[nicoc]

Thank you very much for your patience and such a nice welcome to the forum!