# What are the extrema of f(x)=3x^2 - 12x + 13 on [-oo,oo]?

Oct 17, 2016

$f \left(x\right)$ has a minimum at $x = 2$

#### Explanation:

Before proceeding, note that this is an upwards facing parabola, meaning we can know without further calculation that it will have no maxima, and a single minimum at its vertex. Completing the square would show us that $f \left(x\right) = 3 {\left(x - 2\right)}^{2} + 1$, giving the vertex, and thus the sole minimum, at $x = 2$. Let's see how this would be done with calculus, though.

Any extrema will occur either at a critical point or at an endpoint of the given interval. As our given interval of $\left(- \infty , \infty\right)$ is open, we can ignore the possibility of endpoints, and so we will first identify the critical points of the function, that is, the point at which the derivative of the function is $0$ or does not exist.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 12 x + 13\right) = 6 x - 12$

Setting this equal to $0$, we find a critical point at $x = 2$

$6 x - 12 = 0 \implies x = \frac{12}{6} = 2$

Now, we can either test to see whether it is an extremum (and what type) by checking some values of $f$ around that point, or by using the second derivative test. Let's use the latter.

$\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left(6 x - 12\right) = 6$

As $f ' ' \left(2\right) = 6 > 0$, the second derivative test tells us that $f \left(x\right)$ has a local minimum at $x = 2$

Thus, using $f ' \left(x\right)$ and $f ' ' \left(x\right)$, we find that $f \left(x\right)$ has a minimum at $x = 2$, matching the result we found using algebra.