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

Answer 1

The correct option is A f(-x)

Given equation is $f\left(x\right)=4x^3+3x^2+3x+4$

We have to find $x^{3}f\left(\frac{1}{3}\right)$

Substituting $x=\frac{1}{x}$ in the given equation:

Thus $f\left(\frac{1}{x}\right)=4{\left(\frac{1}{x}\right)}^3+3{\left(\frac{1}{x}\right)}^2+3\left(\frac{1}{x}\right)+4$

$=\frac{4}{x^3}+\frac{3}{x^2}+\frac{3}{x}+4$

Therefore, $x^{3}f\left(\frac{1}{x}\right)$ $=x^3(\frac{4}{x^3}+\frac{3}{x^2}+\frac{3}{x}+4)$

$=4+3x+3x^2+4x^3$

$=4x^3+3x^2+3x+4=f\left(x\right)$

Option D is correct.