
If $f\left( x \right)=-2{{x}^{2}}-5x$ and $g\left( x \right)=3x+2$, how do you find the domain and range of $f\left( x \right),g\left( x \right)$ and $f\left( g\left( x \right) \right)$?
Answer
533.4k+ views
Hint: We first describe the use of a function. From the given function we find the restrictions on $x$ which in turn helps in finding the domain. The whole function is then put under the domain to find the range. We do the same thing for the composite function.
Complete step by step solution:
The basic requirement for a function to have input and evaluate its output. The input is called the domain of the function and the outcome is called the range.
We generally define a function as $f\left( x \right)$. We can also express it as $f:x\to f\left( x \right)$.
Here for the given functions, we have $f\left( x \right)=-2{{x}^{2}}-5x$ and $g\left( x \right)=3x+2$. The power value of 2 and 1 cannot have any restrictions. Therefore, we can use any real value for the $x$.
The domain for the function $f\left( x \right)=-2{{x}^{2}}-5x$ and $g\left( x \right)=3x+2$ is $\mathbb{R}$.
Now we have to find the range for the function.
The linear function $g\left( x \right)=3x+2$ can take any real value as it is a one-one function. The range for $g\left( x \right)=3x+2$ is $\mathbb{R}$.
Now $f\left( x \right)=-2{{x}^{2}}-5x=\dfrac{25}{8}-2{{\left( x+\dfrac{5}{4} \right)}^{2}}$.
We know $\forall x\in \mathbb{R}$, the value of ${{\left( x+\dfrac{5}{4} \right)}^{2}}\ge 0$.
Therefore, we have a maximum value for $f\left( x \right)=-2{{x}^{2}}-5x$ which is $\dfrac{25}{8}$. The range for the function is $\left( -\infty ,\dfrac{25}{8} \right]$.
Now we find the domain and range for $f\left( g\left( x \right) \right)$. It is a composite function where the domain id is determined by $g\left( x \right)=3x+2$ and range is determined by $f\left( x \right)=-2{{x}^{2}}-5x$.
We have $f\left( g\left( x \right) \right)=f\left( 3x+2 \right)=-2{{\left( 3x+2 \right)}^{2}}-5\left( 3x+2 \right)=-18{{x}^{2}}-39x-18$.
Therefore, the domain and the range for the function $f\left( g\left( x \right) \right)$ are $\mathbb{R}$ and $\left( -\infty ,\dfrac{25}{8} \right]$ respectively.
Note: We have to be careful about the equality of the function’s equality for the boundary points. If the boundary points are not in the domain then the function would not be defined at that point.
Complete step by step solution:
The basic requirement for a function to have input and evaluate its output. The input is called the domain of the function and the outcome is called the range.
We generally define a function as $f\left( x \right)$. We can also express it as $f:x\to f\left( x \right)$.
Here for the given functions, we have $f\left( x \right)=-2{{x}^{2}}-5x$ and $g\left( x \right)=3x+2$. The power value of 2 and 1 cannot have any restrictions. Therefore, we can use any real value for the $x$.
The domain for the function $f\left( x \right)=-2{{x}^{2}}-5x$ and $g\left( x \right)=3x+2$ is $\mathbb{R}$.
Now we have to find the range for the function.
The linear function $g\left( x \right)=3x+2$ can take any real value as it is a one-one function. The range for $g\left( x \right)=3x+2$ is $\mathbb{R}$.
Now $f\left( x \right)=-2{{x}^{2}}-5x=\dfrac{25}{8}-2{{\left( x+\dfrac{5}{4} \right)}^{2}}$.
We know $\forall x\in \mathbb{R}$, the value of ${{\left( x+\dfrac{5}{4} \right)}^{2}}\ge 0$.
Therefore, we have a maximum value for $f\left( x \right)=-2{{x}^{2}}-5x$ which is $\dfrac{25}{8}$. The range for the function is $\left( -\infty ,\dfrac{25}{8} \right]$.
Now we find the domain and range for $f\left( g\left( x \right) \right)$. It is a composite function where the domain id is determined by $g\left( x \right)=3x+2$ and range is determined by $f\left( x \right)=-2{{x}^{2}}-5x$.
We have $f\left( g\left( x \right) \right)=f\left( 3x+2 \right)=-2{{\left( 3x+2 \right)}^{2}}-5\left( 3x+2 \right)=-18{{x}^{2}}-39x-18$.
Therefore, the domain and the range for the function $f\left( g\left( x \right) \right)$ are $\mathbb{R}$ and $\left( -\infty ,\dfrac{25}{8} \right]$ respectively.
Note: We have to be careful about the equality of the function’s equality for the boundary points. If the boundary points are not in the domain then the function would not be defined at that point.
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