# Identify the intervals of concavity and the point inflection for f(x)= \sqrt {x^5}

## Question:

Identify the intervals of concavity and the point inflection for {eq}f(x)= \sqrt {x^5} {/eq}

## Inflection point:

The point that, in a continuous function, separates the concave up part from the concave down part, is called the inflection point of the function.The most important thing is that this point must belong to the domain of the function.

Domain {eq}\displaystyle f(x)= \sqrt {x^5}\\ \boxed{\displaystyle D = \{ x \in R \}} {/eq}

To determine the function's concavity and inflection points, we find and analyze the second derivative:

{eq}\displaystyle f(x)= \sqrt {x^5}\\ \displaystyle f'(x)= \frac{5x^4}{7\left(x^5\right)^{\frac{6}{7}}}\\ \displaystyle f''(x) = \frac{5}{7}\cdot \frac{4x^3\left(x^5\right)^{\frac{6}{7}}-\frac{30x^4}{7\left(x^5\right)^{\frac{1}{7}}}x^4}{\left(\left(x^5\right)^{\frac{6}{7}}\right)^2}\\ \displaystyle f''(x) = -\frac{10x^8}{49\left(x^5\right)^{\frac{13}{7}}}\\ \displaystyle \boxed{x=0} \,\, \Longrightarrow \,\, \textrm {critical point of the second derivative} \,\,\, x=0 \, \in \, D {/eq}

To define the sign of the second derivative in each interval, evaluate a point of each interval and verify the sign:

{eq}\left (-\infty,0\right) \,\,\,\, \rightarrow \,\,\,\, f''(-1)=\frac{10}{49} > 0 \,\,\,\,\, \textrm { the second derivative is positive in this interval } \\ \left (0, \infty\right) \,\,\,\, \rightarrow \,\,\,\, f''(-1)=-\frac{10}{49}< 0 \,\,\,\,\, \textrm { the second derivative is negative in this interval } {/eq}

Conclusion (Concavity) {eq}\boxed{ \left (-\infty, 0\right)} \,\, \Longrightarrow \,\, \textrm {concave up}\\ \boxed{ \left (0, \infty\right)} \,\, \Longrightarrow \,\, \textrm {concave down}\\ \boxed{ P_1\left (0,0\right)} \,\, \Longrightarrow \,\, \textrm {inflection point of the function}\\ {/eq} 