The graph of function \(f(x)=\log _e({x}^3+\sqrt[]{{x}^6+1})\) is symmetric about:
\[ f(x) = \log \left[(x)^{3} + \sqrt{(x)^{6} + 1}\right] \] Given the function: \[ f(-x) = \log \left[(-x)^{3} + \sqrt{(-x)^{6} + 1}\right] \] After substituting \( -x \) into the function, you simplify the expression: \[ f(-x) = \log \left[\sqrt{x^{6}+1} - x^{3}\right] \] Then, you rationalize the expression by multiplying the numerator and denominator by \( \left(\sqrt{x^{6}+1} + x^{3}\right) \): \[ f(-x) = \log \left[\frac{\left(\sqrt{x^{6}+1}-x^{3}\right)\left(\sqrt{x^{6}+1}+x^{3}\right)}{\sqrt{x^{6}+1}+x^{3}}\right] \] Simplifying further: \[ f(-x) = \log \left(\frac{1}{x^{3} + \sqrt{x^{6}+1}}\right) \] Which is equivalent to: \[ f(-x) = -\log \left(x^{3} + \sqrt{x^{6}+1}\right) \] And since \( f(x) = \log \left(x^{3} + \sqrt{x^{6}+1}\right) \), we can conclude: \[ f(-x) = -f(x) \] This confirms that the function \( f(x) \) is odd, meaning it is symmetric about the origin.
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