# Find the open intervals where f(x)= -2x^3 + 6x^2 + 168x -8 is concave upward or concave downward....

## Question:

Find the open intervals where {eq}f(x)= -2x^3 + 6x^2 + 168x -8 {/eq} is concave upward or concave downward. Find any inflection points.

## Concavity:

Concavity tells us something about how a curve curves. We can think of upward concavity as a bowl (if the curve were a road and it were to rain, this part would fill with water), and developing the road analogy downward concavity as an overpass (if the curve were a road and it were to rain, then nothing underneath it would get wet). Intervals where the second derivative are positive are intervals of upward concavity, and those where the second derivative is negative have downward concavity. Those points where the concavity flips, i.e. where the second derivative is 0, are called inflection points.

We need the second derivative. We have

{eq}\begin{align*} f' (x) &= \frac{d}{dx} \left( -2x^3 + 6x^2 + 168x -8 \right) \\ &= -6x^2 + 12x + 168 \end{align*} {/eq}

and then

{eq}\begin{align*} f'' (x) &= -12x + 12 \end{align*} {/eq}

So the inflection point is located at

{eq}\begin{align*} -12x + 12 &= 0 \\ 12x &= 12 \\ x &= 1 \end{align*} {/eq}

At this input we have

{eq}\begin{align*} f(1) &= -2 + 6 +168 - 8 \\ &= 164 \end{align*} {/eq}

so the inflection point is

{eq}\boxed{ (1,164) } {/eq}

Note that for {eq}x < 1 {/eq} the second derivative is positive, so for these input values, the function has upward concavity; and for {eq}x < 1 {/eq} the second derivative is negative, so the function has negative concavity. It is also worth noting that once we had the inflection point, all we had to do was recognize that the function is a cubic with a negative leading coefficient, so the curve starts high goes down into a trough, comes back up a hill, then continues down forever. Either way, we come to the same conclusion regarding concavity.

The function is concave upward on the interval {eq}x \in (-\infty, 1) {/eq}

and concave down on the interval {eq}\begin{align*} x \in (1,\infty) \end{align*} {/eq}.