Answer:
[tex]\textsf{Domain}: \quad \left[-\dfrac{1}{2}, \dfrac{1}{2}\right][/tex]
[tex]\textsf{Range}: \quad [0, 1][/tex]
Step-by-step explanation:
Domain: set of all possible input values (x-values)
Range: set of all possible output values (y-values)
Given function:
[tex]f(x)=\sqrt{1-4x^2}[/tex]
As negative numbers don't have real square roots:
[tex]\implies 1-4x^2\geq 0[/tex]
Therefore, to find the domain, solve the inequality.
Subtract 1 from both sides:
[tex]\implies -4x^2\geq -1[/tex]
Divide both sides by -1 (reverse the inequality):
[tex]\implies 4x^2 \leq 1[/tex]
Divide both sides by 4:
[tex]\implies x^2\leq \dfrac{1}{4}[/tex]
[tex]\textsf{For }\:a^n \leq b,\:\:\textsf{if }n\textsf{ is even then }-\sqrt[n]{b} \leq a \leq \sqrt[n]{b}:[/tex]
[tex]\implies -\sqrt[2]{\dfrac{1}{4}} \leq x \leq \sqrt[2]{\dfrac{1}{4}}[/tex]
[tex]\implies -\sqrt{\dfrac{1}{4}} \leq x \leq \sqrt{\dfrac{1}{4}}[/tex]
[tex]\implies -\dfrac{1}{2} \leq x \leq \dfrac{1}{2}[/tex]
Therefore:
[tex]\textsf{Domain}: \quad \left[-\dfrac{1}{2}, \dfrac{1}{2}\right][/tex]
To find the range, input the endpoints of the domain into the function:
[tex]\implies f\left(-\dfrac{1}{2}\right)=\sqrt{1-4\left(-\dfrac{1}{2}\right)^2}=0[/tex]
[tex]\implies f\left(\dfrac{1}{2}\right)=\sqrt{1-4\left(\dfrac{1}{2}\right)^2}=0[/tex]
To find the limit of the range, find the extreme point(s) of the function by differentiating the function and setting it to zero.
[tex]\implies f(x)=(1-4x^2)^{\frac{1}{2}}[/tex]
[tex]\implies f'(x)=\dfrac{1}{2}(1-4x^2)^{-\frac{1}{2}} \cdot -8x[/tex]
[tex]\implies f'(x)=-\dfrac{4x}{\sqrt{1-4x^2}}[/tex]
Setting it to zero and solving for x:
[tex]\implies -\dfrac{4x}{\sqrt{1-4x^2}}=0[/tex]
[tex]\implies -4x=0[/tex]
[tex]\implies x=0[/tex]
Substitute x = 0 into the function:
[tex]\implies f(0)=\sqrt{1-4(0)^2}=1[/tex]
Therefore, the range is [0, 1]
