# Find the minimum value of the parabola f(x) = 3x^2 - 6x + 5.

## Question:

Find the minimum value of the parabola {eq}f(x) = 3x^2 - 6x + 5 {/eq}.

## Minimum of the Function:

In this problem, we will find the minimum value of the given function. We have to use derivative test for finding the maximum and minimum of the function.

Let y = f(x) be any differentiable function.

First of all, we will find the extremum points of the function f(x) by {eq}\displaystyle \frac{{df}}{{dx}} = 0. {/eq}

Suppose that, x = a is the extremum point of the function.

If {eq}\displaystyle {\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} > 0 {/eq}, then we say that x = a is the minimum of the function f(x).

We have to find the maximum value of {eq}\displaystyle f(x) = 3{x^2} - 6x + 5. {/eq}

Differentiating the function f(x) with respect to x we get

{eq}\displaystyle \eqalign{ & \frac{{df}}{{dx}} = \frac{d}{{dx}}\left( {3{x^2} - 6x + 5} \right) \cr & \frac{{df}}{{dx}} = 6x - 6 \cr & {\text{and}}\,\,\,\frac{{{d^2}f}}{{d{x^2}}} = \frac{d}{{dx}}\left( {6x - 6} \right) \cr & \frac{{{d^2}f}}{{d{x^2}}} = 6. \cr} {/eq}

For extreme points, {eq}\displaystyle \frac{{df}}{{dx}} = 0 {/eq}, then

{eq}\displaystyle 6x - 6 = 0 \Rightarrow x = 1 {/eq}

Here, x = 1, is called the extremum point of the function f(x).

Now we will check minimum of the function f(x).

{eq}\displaystyle {\text{At}}\,\,x\, = \,1,\,\,{\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = 1}} = 6 > 0. {/eq}

Hence given function is the minimum at the point x = 1.

Now we will find the minimum value at this point.

{eq}\displaystyle \eqalign{ & f(x) = 3{x^2} - 6x + 5 \cr & f(1) = 3{(1)^2} - 6(1) + 5 \cr & f(1) = 3 - 6 + 5 \cr & f(1) = 2. \cr} {/eq}