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Sagot :
Answer:Rewrite the cartesian equation as a polar equation.
Rewrite the cartesian equation as a polar equation.r
Rewrite the cartesian equation as a polar equation.r2
Rewrite the cartesian equation as a polar equation.r2cos
Rewrite the cartesian equation as a polar equation.r2cos2
Rewrite the cartesian equation as a polar equation.r2cos2(
Rewrite the cartesian equation as a polar equation.r2cos2(θ
Rewrite the cartesian equation as a polar equation.r2cos2(θ)
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6r
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−r
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(θ
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(θ)
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(θ)=
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(θ)=−
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(θ)=−9
Rewrite the cartesian equation as a polar equation.r2cos2(θ)−6rcos(θ)−rsin(θ)=−9Tap to view steps...
The quadratic function which is represented by the graph of a Parabola fails the Horizontal Line Test thus is not a one to one function. Therefore the inverse is not a function unless with restricted domain.
Edit: Adding the solution below to address update in the comments:
y
=
(
x
+
3
)
2
x
=
(
y
+
3
)
2
±
√
x
=
y
+
3
y
=
−
3
±
√
x
=> inverse relation
Restricting the domain: Recall that the domain and the range of the inverse function are the range and the domain of the original function respectively. The range of the original function in this case is
y
≥
0
, that should be the domain of the inverse function so for the inverse function we take the positive square root as follow:
f
−
1
(
x
)
=
−
3
+
√
x
=> this is the inverse function the domain and the range are:
domain:
x
≥
0
range:
y
≥
−
3
This means we need to restrict the domain of the original function to:
x
≥
−
3
in order for it to have an inverse function.
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