If $h(x)=2g(2x^3-3x^2)+g(6x^2-4x^3-3)\forall x\in\mathbb{R}$ and $g''(x)>0\forall x\in\mathbb{R}$. Then find values of $x$ for which $h(x)$ is increasing
solution i try
$h(x)=2g'(2x^3-3x^2)(6x^2-6x)-g'(6x^2-4x^3-3)(12x^2-12x)$
From $g''(x)>0$ implies $g'(x)$ is increasing function.
Not be able to solve ahead Help required
On
Jacky.
$$ h(x) = 2g(2x^3−3x^2)+g(6x^2−4x^3−3) $$ Now let $g(2x^3−3x^2) = g(a)$ and $g(6x^2−4x^3−3)=g(b).$ So the 1st deriv of $h(x)$ is : $$h'(x) = 12 \left[ g'(a)(x^2 - x) + g'(b)(x-x^2) \right] $$ $$ = 12 (x^2-x)\left[ g'(a) - g'(b) \right] $$ Now $h'(x)$ will be positive when $$ x^2-x > 0 \:\: \text{and} \:\: g'(a)>g'(b) $$ $$ \cup $$ $$ x^2-x < 0 \:\: \text{and} \:\: g'(a) < g'(b) $$ Since we know that $g'(x)$ is increasing, then $g'(a) > g'(b)$ is achieved when $a > b$, or $2x^3-3x^2 > 6x^2-4x^3-3$. so now $h'(x)$ will be positive when $$ x^2-x > 0 \:\: \text{and} \:\: 2x^3-3x^3 > 6x^2-4x^3-3 $$ $$ \cup $$ $$ x^2-x < 0 \:\: \text{and} \:\: 2x^3-3x^3 < 6x^2-4x^3-3 $$
Can you continue from here?
\begin{align*} h'(x)&=12x(x-1)[g'(2x^{3}-3x^{2})-g'(6x^{2}-4x^{3}-3)]\\ &=12x(x-1)g''(\eta_{x})(6x^{3}-9x^{2}+3)\\ &=36x(x-1)(x-1)(2x^{2}-x-1)g''(\eta_{x})\\ &=36x(x-1)^{2}(x-1)(2x+1)g''(\eta_{x})\\ &=72x(x-1)^{3}(x+1/2)g''(\eta_{x}), \end{align*} $h'(x)>0$ whenever $x>1$ and $-1/2<x<0$.