Question

which statement is true of the function $f(x) = -sqrt3{x}$? select three options.

• the function is always increasing.
• the function has a domain of all real numbers.
• the function has a range of ${y|-\infty < y < \infty}$.
• the function is a reflection of $y = sqrt3{x}$.
• the function passes through the point $(3, -27)$.

Explanation:

Step1: Analyze "always increasing"

The function \( f(x) = -\sqrt[3]{x} \). The derivative (or by analyzing the cube root function) shows that as \( x \) increases, \( \sqrt[3]{x} \) increases, so \( -\sqrt[3]{x} \) decreases. So "always increasing" is false.

Step2: Analyze domain

For cube root function \( \sqrt[3]{x} \), \( x \) can be any real number (positive, negative, zero). So domain of \( f(x) = -\sqrt[3]{x} \) is all real numbers. This is true.

Step3: Analyze range

Since \( x \) can be any real number, \( \sqrt[3]{x} \) can be any real number, so \( -\sqrt[3]{x} \) can also be any real number. So range is \( \{ y | -\infty < y < \infty \} \). This is true.

Step4: Analyze reflection

The function \( y = \sqrt[3]{x} \) reflected over the x - axis (multiplying by - 1) gives \( y=-\sqrt[3]{x} \). So it is a reflection. This is true.

Step5: Analyze point (3, - 27)

Substitute \( x = 3 \) into \( f(x)=-\sqrt[3]{3}\approx - 1.44
eq - 27 \). So this is false.

Answer:

The function has a domain of all real numbers, The function has a range of \(\{y|-\infty < y < \infty\}\), The function is a reflection of \(y = \sqrt[3]{x}\)