# Find the point of inflection and the interval(s) of concavity of the graph of the function f(x)=...

## Question:

Find the point of inflection and the interval(s) of concavity of the graph of the function {eq}f(x)= x^3 - 3x^2 + 7x. {/eq}

## Concavity and Inflection Points Using Second-Order Derivatives:

Given a continuous function {eq}F(x){/eq}. Its second-order derivative describes the functions concavity such that at any point {eq}x=c{/eq} on the domain of the function, {eq}F(x){/eq}:

• is concave up if {eq}F''(c) > 0{/eq};
• is concave down if {eq}F''(c) < 0{/eq}; and,
• has an inflection point if {eq}F''(c) = 0{/eq}.

Given the function {eq}f(x) = x^3 - 3x^2 + 7x{/eq}.

Solve for the zeros of the second-order derivative of the function to locate the points of inflection:

{eq}\begin{align*} f''(x) &= \frac{d^2}{dx^2} \left( x^3 - 3x^2 + 7x \right) & \text{[Differentiate twice with respect to } x \text{]} \\ &= \frac{d}{dx} \left( 3x^2 - 6x + 7 \right) \\ f''(x) &= 6x - 6 & \text{[Solve for the zeros of } f''(x) \text{]} \\ 0 &= 6x - 6 \\ x &= 1 & \boxed{f(x) \text{ has an inflection point at } x=1. } \end{align*} {/eq}

This inflection point also divides the domain of the function into the intervals {eq}(-\infty, 1) \text{ and } (1,\infty){/eq}.

Determine whether the function is concave up or down on each interval by evaluating {eq}f''(x){/eq} on an arbitrary value from the said interval:

{eq}\begin{align*} &(-\infty, 1) \text{: let } x = 0 \\ &\Rightarrow f''(0) = 6(0) - 6 \\ &\Rightarrow f''(0) = -6 < 0 & \boxed{ f(x) \text{ is concave down on the interval } (-\infty,1). } \\ \\ &(1,\infty) \text{: let } x = 2 \\ &\Rightarrow f''(2) = 6(2) - 6 \\ &\Rightarrow f''(2) = 6 > 0 & \boxed{ f(x) \text{ is concave up on the interval } (-\infty,1). } \end{align*} {/eq}