# What are the extrema of f(x)=4x^2-24x+1?

Mar 17, 2016

The function has a minimum at $x = 3$ where $f \left(3\right) = - 35$

#### Explanation:

$f \left(x\right) = 4 {x}^{2} - 24 x + 1$

The 1st derivative gives us the gradient of the line at a particular point. If this is a stationary point this will be zero.

$f ' \left(x\right) = 8 x - 24 = 0$

$\therefore 8 x = 24$

$x = 3$

To see what type of stationary point we have we can test to see if the 1st derivative is increasing or decreasing. This is given by the sign of the 2nd derivative:

$f ' ' \left(x\right) = 8$

Since this is +ve the 1st derivative must be increasing indicating a minimum for $f \left(x\right)$.

graph{(4x^2-24x+1) [-20, 20, -40, 40]}

Here $f \left(3\right) = 4 \times {3}^{2} - \left(24 \times 3\right) + 1 = - 35$