Question

1. Find the domain and range of the function f(x)=√1-4x².

Answer 1

$f(x) = \sqrt{1 - 4{ x}^{2} }$

$1 - 4x {}^{2} \geqslant 0$

$- \infty \: \: \: \: \: \: \: \: - 0.5 \: \: \: \: \: \: \: \: \: 0.5 \: \: \: \: \: \: \: \: \infty \\ - \: \: \: \: \: \: \: \: \: \: \: \: \: \:0 \: \: \: \: \: + \: \: \: \:0 \: \: \: \: \: -$

$domain \\ [ \: -0.5 \: , \: 0.5 \: ]$

$g(x) = 1 - 4x {}^{2} \\ g'(x) = - 8x \\ g'(x) = 0 \\ x = 0 \\ g(0) = 1 - 4(0) = 1 \\ maximum \: = 1$

$range \\ [ \: 0 \: , \: 1 \: ]$

Answer 2

Answer:

$\textsf{Domain}: \quad \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

$\textsf{Range}: \quad [0, 1]$

Step-by-step explanation:

Domain: set of all possible input values (x-values)

Range: set of all possible output values (y-values)

Given function:

$f(x)=\sqrt{1-4x^2}$

As negative numbers don't have real square roots:

$\implies 1-4x^2\geq 0$

Therefore, to find the domain, solve the inequality.

Subtract 1 from both sides:

$\implies -4x^2\geq -1$

Divide both sides by -1 (reverse the inequality):

$\implies 4x^2 \leq 1$

Divide both sides by 4:

$\implies x^2\leq \dfrac{1}{4}$

$\textsf{For }\:a^n \leq b,\:\:\textsf{if }n\textsf{ is even then }-\sqrt[n]{b} \leq a \leq \sqrt[n]{b}:$

$\implies -\sqrt[2]{\dfrac{1}{4}} \leq x \leq \sqrt[2]{\dfrac{1}{4}}$

$\implies -\sqrt{\dfrac{1}{4}} \leq x \leq \sqrt{\dfrac{1}{4}}$

$\implies -\dfrac{1}{2} \leq x \leq \dfrac{1}{2}$

Therefore:

$\textsf{Domain}: \quad \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

To find the range, input the endpoints of the domain into the function:

$\implies f\left(-\dfrac{1}{2}\right)=\sqrt{1-4\left(-\dfrac{1}{2}\right)^2}=0$

$\implies f\left(\dfrac{1}{2}\right)=\sqrt{1-4\left(\dfrac{1}{2}\right)^2}=0$

To find the limit of the range, find the extreme point(s) of the function by differentiating the function and setting it to zero.

$\implies f(x)=(1-4x^2)^{\frac{1}{2}}$

$\implies f'(x)=\dfrac{1}{2}(1-4x^2)^{-\frac{1}{2}} \cdot -8x$

$\implies f'(x)=-\dfrac{4x}{\sqrt{1-4x^2}}$

Setting it to zero and solving for x:

$\implies -\dfrac{4x}{\sqrt{1-4x^2}}=0$

$\implies -4x=0$

$\implies x=0$

Substitute x = 0 into the function:

$\implies f(0)=\sqrt{1-4(0)^2}=1$

Therefore, the range is [0, 1]