Question

Consider the function f(x) = 3x2 - 12x-4.
a. Determine, without graphing, whether the function has a minimum value or a maximum value.
b. Find the minimum or maximum value and determine where it occurs.
c. Identify the function's domain and its range.

...

a. The function has a minimum or maximum value.

b. The minimum/maximum value is blank. It occurs at x= blank.

c. The domain of f is blank (Type your answer in interval​ notation.)

The range of f is blank (Type your answer in interval​ notation.)

Answer 1

To analyze the function \( f(x) = 3x^2 - 12x - 4 \):

### a. Determine whether the function has a minimum or a maximum value.
The function is a quadratic function of the form \( ax^2 + bx + c \), where \( a = 3 \), \( b = -12 \), and \( c = -4 \).

Since \( a = 3 \) is positive, the parabola opens upwards, which means the function has a **minimum value**.

### b. Find the minimum value and determine where it occurs.
To find the minimum value, we can use the vertex formula for a quadratic function:
\[
x = -\frac{b}{2a}
\]

Plugging in the values of \( a \) and \( b \):
\[
x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2
\]

Now, we need to find the function's value at \( x = 2 \):
\[
f(2) = 3(2)^2 - 12(2) - 4
\]
\[
= 3(4) - 24 - 4
\]
\[
= 12 - 24 - 4
\]
\[
= 12 - 28 = -16
\]

So, the minimum value is \( -16 \) and it occurs at \( x = 2 \).

### c. Identify the function's domain and its range.
- **Domain**: Since \( f(x) \) is a polynomial, its domain is all real numbers:
\[
\text{Domain} = (-\infty, \infty)
\]

- **Range**: Since the minimum value is \( -16 \) and the parabola opens upwards, the range starts from \( -16 \) and goes to positive infinity:
\[
\text{Range} = [-16, \infty)
\]

### Summary of the Answers:
a. The function has a **minimum** value.

b. The minimum value is **-16**. It occurs at \( x = 2 \).

c. The domain of \( f \) is **(-\infty, \infty)**. The range of \( f \) is **[-16, \infty)**.