Question

F(x)=3x^2+30x-1Does the quadratic function f have a minimum value or a maximum value? What is the minimum or maximum value?

Answer 1

To solve this question we will use the first and second derivative criteria.

The first and second derivatives of the given function are:

`\begin{gathered} F^(\prime)(x)=2*3x+30=6x+30, \\ F^(\prime)^(\prime)(x)=6. \end{gathered}`

Since the second derivative is greater than zero for all real numbers, we have that the function reaches a minimum.

Setting F'(x)=0 we get:

`0=6x+30.`

Subtracting 30 from the above equation we get:

`\begin{gathered} 6x+30-30=0-30, \\ 6x=-30. \end{gathered}`

Now, dividing by 6 we get:

`\begin{gathered} (6x)/(6)=-(30)/(6), \\ x=-5. \end{gathered}`

Therefore F(x) reaches a minimum at x=-5.

The minimum value is:

`\begin{gathered} F(-5)=3(-5)^2+30(-5)-1 \\ =3\cdot25-150-1=75-150-1 \\ =-76. \end{gathered}`

Answer: The given function has a minimum value of -76.