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.
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Domain: Since \( f(x) \) is a polynomial, its domain is all real numbers: \[ \text{Domain} = (-\infty, \infty) \]
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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).
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