How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for y=-(-3x+3)^(1/2)?

Sep 29, 2017

The function is convex on the interval $\left(- \infty , 1\right)$

Explanation:

Calculate the first and second derivatives

$y = - \sqrt{3 - 3 x}$

The domain of definition is $I = \left(- \infty , 1\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \cdot \left(- 3\right) {\left(3 - 3 x\right)}^{- \frac{1}{2}} = \frac{3}{2} {\left(3 - 3 x\right)}^{- \frac{1}{2}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{3}{2} \cdot - \frac{3}{2} {\left(3 - 3 x\right)}^{- \frac{3}{2}} = \frac{9}{4} {\left(3 - 3 x\right)}^{- \frac{3}{2}}$

$\forall x \in I$, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 > 0$, the function is convex

graph{-sqrt(3-3x) [-16.02, 16.01, -8.01, 8.01]}