If g(x) = frac{1}{6} f(3 x^2 - 1) + frac{1}{2 | vedclass

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(A) Given $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2)$.Taking the derivative with respect to $x$:$g'(x) = \frac{1}{6} f'(3 x^2 - 1) \cdo

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$\left( \frac{-1}{\sqrt{2}}, 0 \right) \cup \left( \frac{1}{\sqrt{2}}, \infty \right)$

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(A) Given $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2)$.Taking the derivative with respect to $x$:$g'(x) = \frac{1}{6} f'(3 x^2 - 1) \cdot (6x) + \frac{1}{2} f'(1 - x^2) \cdot (-2x)$$g'(x) = x [f'(3 x^2 - 1) - f'(1 - x^2)]$.Since $f''(x) > 0$,$f'(x)$ is an increasing function.For $g(x)$ to be increasing,$g'(x) > 0$.Case $1$: If $x > 0$,then we need $f'(3 x^2 - 1) - f'(1 - x^2) > 0$,which implies $f'(3 x^2 - 1) > f'(1 - x^2)$.Since $f'$ is increasing,this means $3 x^2 - 1 > 1 - x^2$,so $4 x^2 > 2$,or $x^2 > \frac{1}{2}$.Given $x > 0$,this gives $x \in \left( \frac{1}{\sqrt{2}}, \infty \right)$.Case $2$: If $x Since $f'$ is increasing,this means $3 x^2 - 1 Given $x Thus,$g(x)$ is increasing in $\left( -\frac{1}{\sqrt{2}}, 0 \right) \cup \left( \frac{1}{\sqrt{2}}, \infty \right)$.

If $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2), \forall x \in R$,where $f''(x) > 0, \forall x \in R$. Then $g(x)$ is increasing in the interval

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