Determine, without graphing, whether the function f(x)=4x^2-8x+3 has a maximum value or a minimum value and then find the value.

We are asked to determine whether the function has a maximum or a minimum, and then find the value.

The graph of the function is a parabola; since the leading coefficient is positive the parabola opens up and therefore there is a minimum at the vertex.

The vertex has x-coefficient found by: here a=4 and b=-8 so the x-coordinate is . The y-coordinate is.

The minimum value occurs at (1,-1).

(1) Alternatively we can rewrite in vertex form:

where the vertex is at (1,-1); the graph opens up since the leading coefficient is positive so there is a minimum at the vertex.

(2) Using calculus:

so  . The derivative is zero when.

For x<1 the derivative is negative so the function decreases and the derivative is positive for x>1 so the function increases on this interval.

Decreasing then increasing indicates that the critical point at x=1 is a minimum for the function. so there is a minimum at (1,-1). This is the only critical point so it is a global minimum (the function is continuous on the real numbers.)