Question
If $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x)<0$ for $0 \leq x \leq 1$, then(A) $g(x)$ increases in $\left(-\infty, \frac{1}{2}\right)$(B) $g(x)$ increases in $\left(0, \frac{1}{2}\right)$
Step 1
This means that $f(x)$ is concave down on the interval $[0,1]$. Show more…
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If $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x)<0$ for $0 \leq x \leq 1$, then (A) $g(x)$ increases in $\left(-\infty, \frac{1}{2}\right)$ (B) $g(x)$ increases in $\left(0, \frac{1}{2}\right)$ (C) $g(x)$ decreases in $\left(\frac{1}{2}, 1\right)$ (D) $g(x)$ decreases in $\left(\frac{1}{2}, \infty\right)$
If $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x)<0 ; 0 \leq x \leq 1$, show that $g(x)$ increases in $0<x<\frac{1}{2}$ and decreases in $\frac{1}{2}<x<1$
If $f^{\prime}(0)=0$ and $f^{\prime \prime}(x)>0$ for $x \geq 0,$ then $f$ is increasing on $[0, \infty).$
Applications of the Derivative
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