# How do you determine whether the function f(x)= -6 sqrt (x) is concave up or concave down and its intervals?

Jul 28, 2015

Use calculus (the sign of the second derivative) or algebra/precalculus graphing techniques.

#### Explanation:

Calculus

In general, to investigate concavity of the graph of function $f$, we investigate the sign of the second derivative.

$f \left(x\right) = - 6 {x}^{\frac{1}{2}}$

Note first that the doamin of $f$ is $\left[0 , \infty\right)$

$f ' \left(x\right) = - 3 {x}^{- \frac{1}{2}}$

$f ' ' \left(x\right) = \frac{3}{2} {x}^{- \frac{3}{2}} = \frac{3}{2 {\sqrt{x}}^{3}}$

$f ' ' \left(x\right)$ is positive for all real $x$, so it is positive for all $x$ in the domain of $f$.

The graph of $f$ is concave up on $\left(0 , \infty\right)$.

(Intervals of concavity are generally given as open intervals.)

Algebra/Precalculus

The graph of the square root function looks like this:

graph{y = sqrtx [-10, 10, -5, 5]}

That graph is concave down.

Multiplying by $- 6$ reflects the graph across the $x$ asix and stretches it vertically by a factor of $6$:

graph{y = -6sqrtx [-6.41, 25.63, -13.2, 2.82]}

The graph is concave up.