
How do you find the domain and range of $f\left( x \right)=5{{x}^{2}}+2x-1$ ?
Answer
549.9k+ views
Hint: We have been given a quadratic equation in x-variable whose domain and range is to be computed. In order to find the domain of the function, we shall find all the values of x for which the function is defined and in order to find the range of the function, we shall find all the values of $f\left( x \right)$ which comes as output for each and every value of x given as input to the function.
Complete step by step solution:
We can view a function, $f$ as something which takes input $x$ and for a given input, it produces an output which we call $f\left( x \right)$.
Given that, $f\left( x \right)=5{{x}^{2}}+2x-1$
Any quadratic function or any polynomial in general does not have any restrictions to the value which shall be given as input to the function. The quadratic functions are defined for all real numbers which means that any real number can be given as input and substituted in place of x to give the respective output of the function.
Since the coefficient of the 2-degree term, ${{x}^{2}}$ is 5 which is greater than zero, thus, the range of the function is from the vertex of the graph of function to infinity.
The vertex of a quadratic function, $a{{x}^{2}}+bx+c$ is given as $c-\dfrac{{{b}^{2}}}{4a}$.
Thus the vertex of $5{{x}^{2}}+2x-1$ is y is given as:
\[y=-\left( -1 \right)-\dfrac{{{\left( 2 \right)}^{2}}}{4\left( 5 \right)}\]
\[\Rightarrow y=1-\dfrac{4}{20}\]
\[\Rightarrow y=1-0.2\]
\[\Rightarrow y=0.8\]
Hence the range of the function is $\left[ 0.8,\infty \right)$.
Therefore, for the function $f\left( x \right)=5{{x}^{2}}+2x-1$, domain is $x\in \mathbb{R}$ and range is $y\in \left[ 0.8,\infty \right)$.
Note:
The graph of any quadratic equation of the form, $a{{x}^{2}}+bx+c$ is upward opening if $a > 0$ and downward opening if $a<0$. The range of quadratic equations with upward opening parabolic graphs is from the vertex of the graph upto infinity whereas the range of quadratic equations with downward opening parabola graphs is from the vertex of the graph up to negative infinity.
Complete step by step solution:
We can view a function, $f$ as something which takes input $x$ and for a given input, it produces an output which we call $f\left( x \right)$.
Given that, $f\left( x \right)=5{{x}^{2}}+2x-1$
Any quadratic function or any polynomial in general does not have any restrictions to the value which shall be given as input to the function. The quadratic functions are defined for all real numbers which means that any real number can be given as input and substituted in place of x to give the respective output of the function.
Since the coefficient of the 2-degree term, ${{x}^{2}}$ is 5 which is greater than zero, thus, the range of the function is from the vertex of the graph of function to infinity.
The vertex of a quadratic function, $a{{x}^{2}}+bx+c$ is given as $c-\dfrac{{{b}^{2}}}{4a}$.
Thus the vertex of $5{{x}^{2}}+2x-1$ is y is given as:
\[y=-\left( -1 \right)-\dfrac{{{\left( 2 \right)}^{2}}}{4\left( 5 \right)}\]
\[\Rightarrow y=1-\dfrac{4}{20}\]
\[\Rightarrow y=1-0.2\]
\[\Rightarrow y=0.8\]
Hence the range of the function is $\left[ 0.8,\infty \right)$.
Therefore, for the function $f\left( x \right)=5{{x}^{2}}+2x-1$, domain is $x\in \mathbb{R}$ and range is $y\in \left[ 0.8,\infty \right)$.
Note:
The graph of any quadratic equation of the form, $a{{x}^{2}}+bx+c$ is upward opening if $a > 0$ and downward opening if $a<0$. The range of quadratic equations with upward opening parabolic graphs is from the vertex of the graph upto infinity whereas the range of quadratic equations with downward opening parabola graphs is from the vertex of the graph up to negative infinity.
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