If $$g(x)=2f2x3$$−$$3x2+f6x2$$−$$4x3$$−$$3$$,∀x∈R and f′′(x)$$0$$,∀x∈R, then $$g(x)$$ is incresing in

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If $$g(x)=2f2x3$$−$$3x2+f6x2$$−$$4x3$$−$$3$$,∀x∈R and f′′(x)>$$0$$,∀x∈R, then $$g(x)$$ is incresing in

Infinity Learn · Accepted Answer

For $$g(x)$$ to be increasing function g′(x)>$$0$$,∀x Since f′′(x)>$$0$$⇒f′(x) is increasing function  Now, $$g(x)=2f2x3$$−$$3x2+f6x2$$−$$4x3$$−$$3g$$′(x)=26x2$$−$$6xf$$′$$2x3$$−$$3x2+12x$$−$$12x2f$$′$$6x2$$−$$4x3$$−$$3g$$′$$(x)=12x(x$$−$$1)f$$′$$2x3$$−$$3x2$$−f′$$6x2$$−$$4x3$$−$$3$$>$$0$$ case $$1$$:if f′$$2x3$$−$$3x2$$>f′$$6x2$$−$$4x3$$−$$3$$⇒$$2x3$$−$$3x2$$>$$6x2$$−$$4x3$$−$$3$$∵f′$$(x) is increases ⇒$$2x3$$−$$3x2+1$$>$$0$$⇒(x$$−$$1)2(2x+1)>$$0$$  ⇒x>$$-12$$∴g′(x)>$$0$$ ⇒$$x(x$$−$$1)$$>$$0$$ and (2x+1)>$$0$$∴$$g(x)$$ is increasing on −$$12$$,$$0$$∪(1$$,∞$$)  case $$2$$:if f′$$2x3$$−$$3x20$$

$If g (x) = 2 f (2 x^{3} - 3 x^{2}) + f (6 x^{2} - 4 x^{3} - 3), \forall x \in R and f^{''} (x) > 0, \forall x \in R, then g (x) is incresing in$

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If g(x)=2f2x3−3x2+f6x2−4x3−3,∀x∈R and f′′(x)>0,∀x∈R, then g(x) is incresing in

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$If g (x) = 2 f (2 x^{3} - 3 x^{2}) + f (6 x^{2} - 4 x^{3} - 3), \forall x \in R and f^{''} (x) > 0, \forall x \in R, then g (x) is incresing in$