Question

Determine whether each function has a maximum or minimum value. Then find the maximum or minimum value of the function.

$f\left(x\right)=3x^2$

Short answer

The function has minimum value that is 0.

Step-by-step solution

Step 1. Use the concept.

Consider the function $f\left(x\right)=a{x}^2+bx+c,a\ne 0$, the x-coordinate of vertex is$\frac{-b}{2a}$

The graph of $f\left(x\right)=a{x}^2+bx+c,a\ne 0$

• opens up and has a minimum value when $a>0$, and
• opens down and has a maximum value when $a<0$

Step 2. Given Information.

The given function is $f\left(x\right)=3x^2$

Step 3. Solution.

In the function $f\left(x\right)=3x^2$, we have $a=3,b=0,c=0$

Here, $a=3>0$

So, the graph opens up and has a minimum value.

The minimum value of the function is the y-coordinate of the vertex.

Thex-coordinate of the vertex is

$\begin{array}{l}\frac{-b}{2a}\\ =\frac{-0}{2\left(3\right)}\mathrm{..........}\left[a=3,b=0\right]\\ =0\end{array}$

Find they-coordinate of the vertex by evaluating the function for $x=0$.

$f\left(0\right)=3{\left(0\right)}^2=0$

So, the minimum value of function is 0.