f(x)=xe0.5xSolution differentiating we get f.pdf

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ANANDCROCKERKOTA

f(x)=xe0.5x Solution differentiating we get f\'(x)=e^.5x+.5x*e^.5x for f\'(X)=0 .5x=-1 or x=-2 f\"(x)=.5e^.5x+.25x*e^.5x+.5e^.5x putting x=-2 f\"(x)>0 so x=-2 is a minima f(-2)=-2/e so f will decrease from -infinity to -2 and increase from -2 to infinity for f\"(X)=0,x=-4,so -4 is the point of inflection so the fuction will be concave from -4 to infinity as f\'\'(x)>0.

f(x)=xe0.5x
Solution
differentiating we get f'(x)=e^.5x+.5x*e^.5x for f'(X)=0 .5x=-1 or x=-2
f"(x)=.5e^.5x+.25x*e^.5x+.5e^.5x putting x=-2 f"(x)>0 so x=-2 is a minima f(-2)=-2/e so f will
decrease from -infinity to -2 and increase from -2 to infinity for f"(X)=0,x=-4,so -4 is the point
of inflection so the fuction will be concave from -4 to infinity as f''(x)>0

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f(x)=xe0.5xSolution differentiating we get f.pdf

1. f(x)=xe0.5x Solution differentiating we get f'(x)=e^.5x+.5x*e^.5x for f'(X)=0 .5x=-1 or x=-2 f"(x)=.5e^.5x+.25x*e^.5x+.5e^.5x putting x=-2 f"(x)>0 so x=-2 is a minima f(-2)=-2/e so f will decrease from -infinity to -2 and increase from -2 to infinity for f"(X)=0,x=-4,so -4 is the point of inflection so the fuction will be concave from -4 to infinity as f''(x)>0