Saturday, July 29, 2017

Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 56

Suppose that all of the functions are twice differentiable and the second derivatives are never 0.

a.) If $f$ and $g$ are concave upward on interval $I$, prove that $f+g$ is concave upward on $I$.
b.) Prove that $g(x) = \left[ f(x) \right]^2$ is concave upward on $I$. Suppose that $f$ is positive and concave upward on $I$.


a.) If both $f$ and $g$ has upward concavity, $f''(x) > 0$ and $g''(x) > 0$ on $I$.
Then, $(f+g)'' = (f'+g')' = f''+g'' > 0$ on $I$.
Therefore, $(f+g)$ has upward concavity at $I$.


b.) If $f$ is positive and has upward concavity on $I$, then $f(x) > 0$ and $f''(x) > 0$

So,

$
\begin{equation}
\begin{aligned}
g'(x) &= 2 f(x) f'(x)\\
\\
g''(x) &= 2 \left[ f(x) f''(x) + f'(x) f'(x) \right]\\
\\
g''(x) &= 2 f(x) f''(x) + 2[f'(x)]^2
\end{aligned}
\end{equation}
$


$f(x)$ and $f''(x)$ are all positive, therefore $g(x)$ has upward concavity at interval $I$.

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