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# Algebra

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The function \$f(x)\$ is defined only on domain \$[-1,2]\$, and is defined on this domain by the formula
\$\$f(x) = 2x^2-8x+1.\$\$
What is the range of \$f(x)\$? Express your answer as an interval or as a union of intervals.

Apr 15, 2024

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We can find the range of the function f(x)=2x2−8x+1 by completing the square and analyzing its extreme points.

Completing the Square:

We can rewrite the function as:

f(x)=2(x2−4x)+1

To complete the square, we take half of the coefficient of our x term, square it, and add it to both sides of the equation:

Half of the coefficient of our x term is -4 / 2 = -2.

Squaring -2 gives us 4.

Therefore:

f(x)=2(x2−4x+4)+1−8

f(x)=2(x−2)2−7

Finding the Vertex and Analyzing Extremes:

The expression inside the parentheses is squared, so it's always non-negative (0 or greater). Multiplying by 2 doesn't change this property. Therefore, the minimum value of f(x) is achieved when the squared term is 0, which happens when x=2. In this case, f(2)=2(2−2)2−7=−7.

Since the coefficient of the squared term (2) is positive, the parabola opens upwards. This means the function has a minimum point at x=2 and increases as we move away from it (either to the left or right on the interval).

Range of the Function:

The function is defined only on the interval [−1,2]. We saw that the minimum value within this interval is f(2)=−7. As x approaches either end of the interval, the function increases without bound because the squared term keeps getting larger. Therefore, the range of f(x) is the single interval:

(−7,∞)​

Apr 15, 2024