If f(x) = 4x^3 + 3x^2 + 3x + 4,then x^3 fleft | vedclass

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(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$.To find $x^3 f\left( \frac{1}{x} \right)$,first substitute $\frac{1}{x}$ for $x$ in the function $f(x)$:$f\lef

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$f(x)$

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(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$.To find $x^3 f\left( \frac{1}{x} \right)$,first substitute $\frac{1}{x}$ for $x$ in the function $f(x)$:$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$$f\left( \frac{1}{x} \right) = \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4$Now,multiply this expression by $x^3$:$x^3 f\left( \frac{1}{x} \right) = x^3 \left( \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4 \right)$$x^3 f\left( \frac{1}{x} \right) = 4 + 3x + 3x^2 + 4x^3$Rearranging the terms,we get:$x^3 f\left( \frac{1}{x} \right) = 4x^3 + 3x^2 + 3x + 4 = f(x)$Therefore,the correct option is $D$.

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

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