If x is a real number, then the minimum value of the function f(x) = 3x² - 2x + 8 is?
Understand the Problem
The question is asking to find the minimum value of the quadratic function f(x) = 3x² - 2x + 8. To solve this, we will calculate the vertex of the parabola represented by this function, as the vertex gives us the minimum value when the parabola opens upwards (which it does since the coefficient of x² is positive).
Answer
The minimum value is $\frac{23}{3}$.
Answer for screen readers
The minimum value of the function $f(x) = 3x^2 - 2x + 8$ is $\frac{23}{3}$.
Steps to Solve
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Identify vertex formula To find the minimum value of the quadratic function $f(x) = ax^2 + bx + c$, we use the vertex formula $x = -\frac{b}{2a}$. In this case, $a = 3$ and $b = -2$.
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Calculate x-coordinate of vertex Using the formula: $$ x = -\frac{-2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3} $$
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Evaluate the function at the vertex Now substitute $x = \frac{1}{3}$ back into the function: $$ f\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) + 8 $$
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Perform calculations Calculate each term:
- First term: $3\left(\frac{1}{3}\right)^2 = 3 \cdot \frac{1}{9} = \frac{1}{3}$
- Second term: $-2\left(\frac{1}{3}\right) = -\frac{2}{3}$
- Combine with the constant term: $$ f\left(\frac{1}{3}\right) = \frac{1}{3} - \frac{2}{3} + 8 = -\frac{1}{3} + 8 = 8 - \frac{1}{3} = \frac{24}{3} - \frac{1}{3} = \frac{23}{3} $$
The minimum value of the function $f(x) = 3x^2 - 2x + 8$ is $\frac{23}{3}$.
More Information
The minimum value occurs at the vertex of the parabola, which is the point where the function reaches its lowest value. Since the parabola opens upwards (as the coefficient of $x^2$ is positive), the vertex represents this minimum.
Tips
- Forgetting to apply the vertex formula correctly.
- Miscalculating the evaluation of the function at the vertex.
- Not simplifying fractions properly.
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