Range of exponential functions
up vote
0
down vote
favorite
I have to find the range of $4^{sin(x)}+ 2^{sin(x)+3}=2^{2sin(x)}+ 8cdot2^{sin(x)}$.
Let's take $y=2^{sin x}$, so we rewrite the equation as $y^2+8y=0$.
The range of this function is $[-16,+infty) $, however, I get $2^{sin x}=-16$ and that's there I'm stuck.
I have a similar problem with $9^x+12cdot3^x+27$. I rewrite it as $x^2+12x+27$. The range is $[-9, infty)$ and I don't know how to apply this to the actual problem.
calculus algebra-precalculus
asked Nov 16 at 1:37
Lowkey
467
add a comment |
up vote
0
down vote
favorite
I have to find the range of $4^{sin(x)}+ 2^{sin(x)+3}=2^{2sin(x)}+ 8cdot2^{sin(x)}$.
Let's take $y=2^{sin x}$, so we rewrite the equation as $y^2+8y=0$.
The range of this function is $[-16,+infty) $, however, I get $2^{sin x}=-16$ and that's there I'm stuck.
I have a similar problem with $9^x+12cdot3^x+27$. I rewrite it as $x^2+12x+27$. The range is $[-9, infty)$ and I don't know how to apply this to the actual problem.
calculus algebra-precalculus
asked Nov 16 at 1:37
Lowkey
467
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to find the range of $4^{sin(x)}+ 2^{sin(x)+3}=2^{2sin(x)}+ 8cdot2^{sin(x)}$.
Let's take $y=2^{sin x}$, so we rewrite the equation as $y^2+8y=0$.
The range of this function is $[-16,+infty) $, however, I get $2^{sin x}=-16$ and that's there I'm stuck.
I have a similar problem with $9^x+12cdot3^x+27$. I rewrite it as $x^2+12x+27$. The range is $[-9, infty)$ and I don't know how to apply this to the actual problem.
calculus algebra-precalculus
asked Nov 16 at 1:37
Lowkey
467
I have to find the range of $4^{sin(x)}+ 2^{sin(x)+3}=2^{2sin(x)}+ 8cdot2^{sin(x)}$.
Let's take $y=2^{sin x}$, so we rewrite the equation as $y^2+8y=0$.
The range of this function is $[-16,+infty) $, however, I get $2^{sin x}=-16$ and that's there I'm stuck.
I have a similar problem with $9^x+12cdot3^x+27$. I rewrite it as $x^2+12x+27$. The range is $[-9, infty)$ and I don't know how to apply this to the actual problem.
calculus algebra-precalculus
calculus algebra-precalculus
asked Nov 16 at 1:37
Lowkey
467
asked Nov 16 at 1:37
Lowkey
467
edited Nov 16 at 1:44
asked Nov 16 at 1:37
Lowkey
467
asked Nov 16 at 1:37
Lowkey
467
asked Nov 16 at 1:37
Lowkey
467
467
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your way of computing the range by using quadratic functions is correct, but you should take care of the domain once you change the variable from $x$ to $y$ because $y=2^{sin(x)}$ can only assume values on $[2^{-1},2]$. Try to decide the range of $y^2+8y$ with domain restricted to $[2^{-1},2]$.
answered Nov 16 at 1:46
Apocalypse
1065
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your way of computing the range by using quadratic functions is correct, but you should take care of the domain once you change the variable from $x$ to $y$ because $y=2^{sin(x)}$ can only assume values on $[2^{-1},2]$. Try to decide the range of $y^2+8y$ with domain restricted to $[2^{-1},2]$.
answered Nov 16 at 1:46
Apocalypse
1065
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
add a comment |
up vote
1
down vote
accepted
Your way of computing the range by using quadratic functions is correct, but you should take care of the domain once you change the variable from $x$ to $y$ because $y=2^{sin(x)}$ can only assume values on $[2^{-1},2]$. Try to decide the range of $y^2+8y$ with domain restricted to $[2^{-1},2]$.
answered Nov 16 at 1:46
Apocalypse
1065
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your way of computing the range by using quadratic functions is correct, but you should take care of the domain once you change the variable from $x$ to $y$ because $y=2^{sin(x)}$ can only assume values on $[2^{-1},2]$. Try to decide the range of $y^2+8y$ with domain restricted to $[2^{-1},2]$.
answered Nov 16 at 1:46
Apocalypse
1065
Your way of computing the range by using quadratic functions is correct, but you should take care of the domain once you change the variable from $x$ to $y$ because $y=2^{sin(x)}$ can only assume values on $[2^{-1},2]$. Try to decide the range of $y^2+8y$ with domain restricted to $[2^{-1},2]$.
answered Nov 16 at 1:46
Apocalypse
1065
answered Nov 16 at 1:46
Apocalypse
1065
answered Nov 16 at 1:46
Apocalypse
1065
answered Nov 16 at 1:46
Apocalypse
1065
1065
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
add a comment |
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
So I just calculate $y^2+8y$ on 1/2 and 2?
– Lowkey
Nov 16 at 1:54
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
Make sure you understand what will happen when you change variables from $x$ to $y=f(x)$: the domain of the new variable $y$ need to be the range of the function $f(x)$. That's exactly where you get stuck, because $2^{sin(x)}$ can never assume value -16.
– Apocalypse
Nov 16 at 1:57
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
If I understand correctly, since the range of $2^(sin^(x))$ is $[1/2,2]$, and y is monotonic and increasing in $[1/2, 2]$, f(1/2) and f(2) will be the minimum and maximum.
– Lowkey
Nov 16 at 2:07
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
Exactly, by the way it should be “the range of $2^{sin(x)}$ is [1/2,2]”, the domain of $2^{sin(x)}$ is still the domain of $sin(x)$ which is $mathbb{R}$, but I think you understand it.
– Apocalypse
Nov 16 at 2:09
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
I can apply this to the second problem, right? the range of $3^x$ is $(0, ∞)$, and f is increasing in that interval, so f(0)=27 will be the minimum. Thanks alot!
– Lowkey
Nov 16 at 2:17
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000595%2frange-of-exponential-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
