# What are the extrema of f(x)=x^2 - 6x + 11  on x in[1,6]?

Nov 5, 2015

$\left(3 , 2\right)$ is a minimum.
$\left(1 , 6\right) \mathmr{and} \left(6 , 11\right)$ are maxima.

#### Explanation:

Relative extrema occur when $f ' \left(x\right) = 0$.
That is, when $2 x - 6 = 0$.
ie when $x = 3$.

To check if $x = 3$ is a relative minimum or maximum, we observe that $f ' ' \left(3\right) > 0$ and so $\implies x = 3$ is a relative minimum,
that is, $\left(3 , f \left(3\right)\right) = \left(3 , 2\right)$ is a relative minimum and also an absolute minimum since it is a quadratic function.

Since $f \left(1\right) = 6 \mathmr{and} f \left(6\right) = 11$, it implies that $\left(1 , 6\right) \mathmr{and} \left(6 , 11\right)$ are absolute maxima on the interval $\left[1 , 6\right]$.

graph{x^2-6x+11 [-3.58, 21.73, -0.37, 12.29]}