F(x)=(x+1)^0.5, h(x)=x^2+3, what is the range of f(h(x))?

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In summary, the range of f(h(x)) is [2, infinity) because the range of h(x) is [3, infinity) and when plugging in the upper bound of [3, infinity), f(h(x)) gives 2.
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Darkmisc
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Homework Statement
f(x)=(x+1)^0.5, h(x)=x^2+3, what is the range of f(h(x))?
Relevant Equations
f(x)=(x+1)^0.5, h(x)=x^2+3
Hi everyone

I have solutions to this problem, but still don't understand them.

I thought the range of h(x) was [3, \infty). Wouldn't this make the domain of f(h(x)) also [3, infinity)?

If this is correct, shouldn't the range of f(h(x)) be [13^0.5, infinity)?

Below is a graph of f(h(x)) using the assumption that the range of h(x) is [3, infinity)

1667269567008.png


The solutions say
1667269818285.png


But the domain of f(h(x)) would need to include 0 for this to be true.
1667269869710.png
Thanks

1667269286916.png

1667269305539.png


1667269331919.png
 

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  • #2
Darkmisc said:
Homework Statement:: f(x)=(x+1)^0.5, h(x)=x^2+3, what is the range of f(h(x))?
Relevant Equations:: f(x)=(x+1)^0.5, h(x)=x^2+3

Hi everyone

I have solutions to this problem, but still don't understand them.

I thought the range of h(x) was [3, \infty). Wouldn't this make the domain of f(h(x)) also [3, infinity)?

If this is correct, shouldn't the range of f(h(x)) be [13^0.5, infinity)?

Below is a graph of f(h(x)) using the assumption that the range of h(x) is [3, infinity)

View attachment 316487

The solutions say View attachment 316489

But the domain of f(h(x)) would need to include 0 for this to be true.
View attachment 316490Thanks

View attachment 316483
View attachment 316484

View attachment 316485
What is the range of h(x)? How does this compare with the domain of f(x)?

-Dan
 
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  • #3
topsquark said:
What is the range of h(x)? How does this compare with the domain of f(x)?

-Dan
Thanks.

range of h(x) is [3, infinity).

If I take [3, infinity) as the domain of f(x), the range becomes [2, infinity), which is the answer from the solutions.

I got the wrong answer by using [3, infinity) as the domain of f(h(x)), although I don't really understand why it is wrong.
 
  • #4
Darkmisc said:
range of h(x) is [3, infinity).

If I take [3, infinity) as the domain of f(x), the range becomes [2, infinity), which is the answer from the solutions.
That's right, because ##[2, \infty)## is the range of ##h##.
Darkmisc said:
I got the wrong answer by using [3, infinity) as the domain of f(h(x)), although I don't really understand why it is wrong.
It's wrong, because ##f(h(x))## is well-defined for all ##x \in \mathbb R##. Just write it down:
$$f(h(x)) = \sqrt{x^2 + 4}$$If I gave that function to a student and asked what was its domain, they wouldn't say ##[3, \infty)##. Why would they?
 
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  • #5
Darkmisc said:
I got the wrong answer by using [3, infinity) as the domain of f(h(x)), although I don't really understand why it is wrong.
You already took care of h(x) when you figured out that its range was ##[3,\infty)##. Now you want the range of f(z), NOT f(h(x)), for z in the domain of ##[3,\infty)##. Plugging 3 into f should give you 2.
 
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FAQ: F(x)=(x+1)^0.5, h(x)=x^2+3, what is the range of f(h(x))?

1. What is the function F(x)?

F(x) is a square root function where the input value (x) is added by 1 and then squared.

2. What is the function h(x)?

h(x) is a quadratic function where the input value (x) is squared and then added by 3.

3. What is the meaning of f(h(x))?

f(h(x)) means that the output of h(x) is used as the input for f(x). In other words, the output of h(x) is plugged into f(x) to get the final result.

4. How do you find the range of f(h(x))?

To find the range of f(h(x)), we need to first find the range of h(x) and then use those values as the input for f(x). The resulting values will be the range of f(h(x)).

5. Can the range of f(h(x)) be negative?

Yes, the range of f(h(x)) can be negative if the range of h(x) includes negative values and those values are used as the input for f(x).

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