We are asked to determine whether the function`f(x)=4x^2-8x+3` 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`x=(-b)/(2a)`: here a=4 and b=-8 so the x-coordinate is `x=8/(8)=1`. The y-coordinate is`f(1)=-1`.
The minimum value occurs at (1,-1).
(1) Alternatively we can rewrite in vertex form:
`f(x)=4(x^2-2x)+3=4(x^2-2x+1)+3-4=4(x-1)^2-1` 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:
`f(x)=4x^2-8x+3` so `f'(x)=8x-8`. The derivative is zero when`x=1`.
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.`f(1)=-1` 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.)