For what values of x is the function f(x) = x | vedclass

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(C) To determine where the function $f(x) = x^9 + 3x^7 + 64$ is strictly increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^9 + 3x^7 +

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All real values of $x$

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(C) To determine where the function $f(x) = x^9 + 3x^7 + 64$ is strictly increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^9 + 3x^7 + 64) = 9x^8 + 21x^6$.Since $x^8 \geq 0$ and $x^6 \geq 0$ for all real $x$,it follows that $9x^8 + 21x^6 \geq 0$ for all $x \in \mathbb{R}$.Specifically,$f'(x) > 0$ for all $x 
eq 0$,and $f'(0) = 0$.Since the derivative is non-negative for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly increasing for all real values of $x$.

For what values of $x$ is the function $f(x) = x^9 + 3x^7 + 64$ strictly increasing?

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