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# Is f(x)=(x-3)^3-x+15 concave or convex at x=3?

Aug 14, 2017

Neither. $x = 3$ is an inflection point.

#### Explanation:

You have to find the second derivative to answer this question.

I would use the chain rule to differentiate ${\left(x - 3\right)}^{3}$. Let $u = x - 3$ and $y = {u}^{3}$. Then $y ' = 3 {u}^{2} \cdot u ' = 3 {u}^{2} \cdot 1 = 3 {\left(x - 3\right)}^{2}$

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

Now it's easy enough to expand to find the second derivative using the power rule.

$f ' \left(x\right) = 3 \left({x}^{2} - 6 x - 9\right) - 1$

$f ' \left(x\right) = 3 {x}^{2} - 18 x - 27 - 1$

$f ' \left(x\right) = 3 {x}^{2} - 18 x - 28$

Now find the second derivative.

$f ' ' \left(x\right) = 6 x - 18$

We now test the sign of the second derivative at $x = 3$ under the knowledge that if:

•f''(x) > 0, at $x = a$ then $f \left(x\right)$ is concave at $x = a$

•f''(x) < 0, at $x = a$ then $f \left(x\right)$ is convex at $x = a$

We have:

$f ' ' \left(3\right) = 6 \left(3\right) - 18 = 18 - 18 = 0$

So $x = 3$ is a point of inflection, so the graph is neither concave up nor concave down at that point. Instead its changing between the two.

Hopefully this helps!