Question

If $f\left(x\right)=4x^3+3x^2+3x+4$, then $x^{3}f\left(\frac{1}{x}\right)$ is equal to:

• A. $f\left(-x\right)$
• B. $\frac{1}{f\left(x\right)}$
• C. ${\left[f\left(\frac{1}{x}\right)\right]}^2$
• D. $f\left(x\right)$

Answer 1

The correct option is D

$f\left(x\right)$

Explanation for the correct option:

Step 1: Find $f\left(\frac{1}{x}\right)$.

$\begin{array}{rcl}f\left(x\right)& =& 4x^3+3x^2+3x+4\\ & \Rightarrow & f\left(\frac{1}{x}\right)=4\left(\frac{1}{x^3}\right)+3\left(\frac{1}{x^2}\right)+3\left(\frac{1}{x}\right)+4\\ & =& \frac{4}{x^3}+\frac{3}{x^2}+\frac{3}{x}+4\end{array}$

Step 2: Find $x^{3}f\left(\frac{1}{x}\right)$.

$\begin{array}{rcl}{x}^{3}f\left(\frac{1}{x}\right)& =& x^3\left[\frac{4}{x^3}+\frac{3}{x^2}+\frac{3}{x}+4\right]\\ & =& 4+3x+3x^2+4x^3\\ & =& f\left(x\right)\end{array}$

Hence, option D is correct.