Question

f"(x)>0 and g(x)=f(2-x)+f(4+x).
then g(x) is increasing in ?

Answer 1

@hartnn

Answer 2

\[\begin{align*}
g(x)&=f(2-x)+f(4+x)\\
\Rightarrow g'(x)&=-f'(2-x)+f'(4+x)&\text{(by the chain rule)}
\end{align*}\]
\(g(x)\) is increasing for values of \(x\) over which \(g'(x)>0\). Are you given that \(f'(x)>0\) or that \(f''(x)>0\)?

Answer 3

ya f"(0)>0 is given

Answer 4

@SithsAndGiggles

Answer 5

g'(x) = -f'(2-x) + f'(4+x)
g'(x) > 0 when -f'(2-x) + f'(4+x) > 0 or
f'(4+x) > f'(2-x)
f''(x) > 0 implies f'(x) is increasing as x increases.
Therefore, f'(4+x) > f'(2-x) when 4+x > 2-x
2x > -2
x > -1

g(x) will be increasing when x > -1

Answer 6

@random321 go for the simpler, correct explanation :)

Answer 7

thank you @ranga and @SithsAndGiggles for the help