Question

Which of the following functions is concave down over the set of positive real numbers:

• A. $f\left(x\right)=2^x$
• B. $f\left(x\right)=x^{\frac{1}{2}}$
• C. $f\left(x\right)=3x+2$
• D. $f\left(x\right)=x^3$

Answer 1

The correct option is B $f\left(x\right)=x^{\frac{1}{2}}$

A function $f\left(x\right)$ is concave down if $f^{\prime \prime }\left(x\right)<0.$
Now, checking options:
$\left(A\right)f^{\prime \prime }\left(x\right)=(\ln 2{)}^2{2}^x$
$\left(B\right)f^{\prime \prime }\left(x\right)=-\frac{1}{4\sqrt{x^3}}$
$\left(C\right)f^{\prime \prime }\left(x\right)=0$
$\left(D\right)f^{\prime \prime }\left(x\right)=6x$
Only $\left(B\right)$ has a negative second derivative for all positive real numbers.