Let's analyze the function \( f(x) = -3x^2 + 18x - 6 \) step by step.
a. Determine whether the function has a minimum or maximum value.
The function is a quadratic function of the form \( ax^2 + bx + c \). In this case, \( a = -3 \), which is less than 0. Since the coefficient of \( x^2 \) (i.e., \( a \)) is negative, the parabola opens downwards. Therefore, the function has a maximum value.
b. Find the maximum value and determine where it occurs.
To find the \( x \)-coordinate of the vertex (which gives the maximum value for downward-opening parabolas), we can use the formula: \[ x = -\frac{b}{2a} \] Here, \( b = 18 \) and \( a = -3 \):
\[ x = -\frac{18}{2(-3)} = -\frac{18}{-6} = 3 \]
Next, we substitute \( x = 3 \) back into the function to find the maximum value:
\[ f(3) = -3(3)^2 + 18(3) - 6 \] \[ = -3(9) + 54 - 6 \] \[ = -27 + 54 - 6 \] \[ = 21 \]
So, the maximum value is 21, and it occurs at \( x = 3 \).
c. Identify the function's domain and its range.
The domain of a quadratic function is all real numbers, since there are no restrictions on \( x \). In interval notation, the domain is: \[ (-\infty, \infty) \]
The range of the function can be determined from the maximum value found earlier. Since the function opens downwards (and hence has a maximum), the range starts from negative infinity up to the maximum value of 21. Therefore, the range is: \[ (-\infty, 21] \]
Final Answers:
a. The function has a maximum value.
b. The maximum value is 21. It occurs at \( x = 3 \).
c. The domain of \( f \) is \( (-\infty, \infty) \).
The range of \( f \) is \( (-\infty, 21] \).
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