Functions within functions
$begingroup$
Am I on the right track? I am learning about relations and functions. Does the domain restriction make it so I have to have my answer between 0 and 17?
Let $f(x) = sqrt{17x-x^2}$ and $g(x) = 3x+4$, defined over a domain of $[0,17]$. Determine $f(g(x))$ and $g(f(x))$.
$(f(g(x)) =$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2+16)}$
$= sqrt{51x+68-9x^2-16)}$
$= sqrt{9x^2-27x-52}$
functions
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
$endgroup$
add a comment |
$begingroup$
Am I on the right track? I am learning about relations and functions. Does the domain restriction make it so I have to have my answer between 0 and 17?
Let $f(x) = sqrt{17x-x^2}$ and $g(x) = 3x+4$, defined over a domain of $[0,17]$. Determine $f(g(x))$ and $g(f(x))$.
$(f(g(x)) =$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2+16)}$
$= sqrt{51x+68-9x^2-16)}$
$= sqrt{9x^2-27x-52}$
functions
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
$endgroup$
add a comment |
$begingroup$
Am I on the right track? I am learning about relations and functions. Does the domain restriction make it so I have to have my answer between 0 and 17?
Let $f(x) = sqrt{17x-x^2}$ and $g(x) = 3x+4$, defined over a domain of $[0,17]$. Determine $f(g(x))$ and $g(f(x))$.
$(f(g(x)) =$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2+16)}$
$= sqrt{51x+68-9x^2-16)}$
$= sqrt{9x^2-27x-52}$
functions
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
$endgroup$
Am I on the right track? I am learning about relations and functions. Does the domain restriction make it so I have to have my answer between 0 and 17?
Let $f(x) = sqrt{17x-x^2}$ and $g(x) = 3x+4$, defined over a domain of $[0,17]$. Determine $f(g(x))$ and $g(f(x))$.
$(f(g(x)) =$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2+16)}$
$= sqrt{51x+68-9x^2-16)}$
$= sqrt{9x^2-27x-52}$
functions
functions
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
asked Nov 29 '18 at 14:55
Arthur GreenArthur Green
796
796
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2color{blue}{+24x}+16)}$
You forgot to write the term $color{blue}{24x}$ from $(a+b)^2=a^2color{blue}{+2ab}+b^2$
$= sqrt{51x+68-9x^2-16color{blue}{-24x}}$
$= sqrt{color{red}{9x^2-27x-52}}$
What happens to the signs here?
$$51x+68-9x^2-16-24x=-9x^2+27x+52 ne color{red}{9x^2-27x-52}$$
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
$endgroup$
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
2
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2color{blue}{+24x}+16)}$
You forgot to write the term $color{blue}{24x}$ from $(a+b)^2=a^2color{blue}{+2ab}+b^2$
$= sqrt{51x+68-9x^2-16color{blue}{-24x}}$
$= sqrt{color{red}{9x^2-27x-52}}$
What happens to the signs here?
$$51x+68-9x^2-16-24x=-9x^2+27x+52 ne color{red}{9x^2-27x-52}$$
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
$endgroup$
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
2
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
add a comment |
$begingroup$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2color{blue}{+24x}+16)}$
You forgot to write the term $color{blue}{24x}$ from $(a+b)^2=a^2color{blue}{+2ab}+b^2$
$= sqrt{51x+68-9x^2-16color{blue}{-24x}}$
$= sqrt{color{red}{9x^2-27x-52}}$
What happens to the signs here?
$$51x+68-9x^2-16-24x=-9x^2+27x+52 ne color{red}{9x^2-27x-52}$$
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
$endgroup$
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
2
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
add a comment |
$begingroup$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2color{blue}{+24x}+16)}$
You forgot to write the term $color{blue}{24x}$ from $(a+b)^2=a^2color{blue}{+2ab}+b^2$
$= sqrt{51x+68-9x^2-16color{blue}{-24x}}$
$= sqrt{color{red}{9x^2-27x-52}}$
What happens to the signs here?
$$51x+68-9x^2-16-24x=-9x^2+27x+52 ne color{red}{9x^2-27x-52}$$
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
$endgroup$
$= sqrt{17(3x+4)-(3x+4)^2}$
$= sqrt{(51x+68)-(9x^2color{blue}{+24x}+16)}$
You forgot to write the term $color{blue}{24x}$ from $(a+b)^2=a^2color{blue}{+2ab}+b^2$
$= sqrt{51x+68-9x^2-16color{blue}{-24x}}$
$= sqrt{color{red}{9x^2-27x-52}}$
What happens to the signs here?
$$51x+68-9x^2-16-24x=-9x^2+27x+52 ne color{red}{9x^2-27x-52}$$
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
answered Nov 29 '18 at 15:01
StackTDStackTD
22.7k2050
22.7k2050
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
2
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
add a comment |
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
2
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
$begingroup$
I wanted to have the $x^2$ term positive so I multipied by negative 1.
$endgroup$
– Arthur Green
Nov 29 '18 at 15:32
2
2
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
$begingroup$
But you're not allowed to do that; you're changing the function... Note that, for example, $f(x)=x$ and $g(x)=-x$ are different functions!
$endgroup$
– StackTD
Nov 29 '18 at 15:33
add a comment |
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