Problem 36 For \(G(x)=x^{3}-8 x^{2}+12 x,\)... [FREE SOLUTION]

Chapter 8: Problem 36

For \(G(x)=x^{3}-8 x^{2}+12 x,\) find all \(x\) -values for which \(G(x) \geq 0\).

Short Answer

Expert verified

x is in [0, 2] or [6, +鈭�).

Step by step solution

Factorize the polynomial

First, factorize the polynomial function. Observe that the polynomial can be rewritten by factoring out an x: \[G(x) = x^3 - 8x^2 + 12x = x(x^2 - 8x + 12).\] Next, factor the quadratic expression: \[x^2 - 8x + 12 = (x-2)(x-6).\] Substituting back, we get \[G(x) = x(x-2)(x-6).\]

Identify the critical points

Set the factors equal to zero to find the critical points: \[x(x-2)(x-6) = 0.\] This gives the solutions \[x = 0, x = 2, x = 6.\] These are the points where \(G(x)\) equals zero.

Test intervals between critical points

Determine the sign of \(G(x)\) in the intervals created by the critical points. The intervals are \((-\infty, 0)\), \((0, 2)\), \((2, 6)\), and \((6, +\infty)\). Choose test points in each interval (e.g., -1, 1, 3, and 7) and substitute into \(G(x)\): For \(x = -1\): \((-1)(-3)(-7) = -21 < 0\), so \(G(x)\) is negative in \((-\infty, 0)\). For \(x = 1\): \(1(-1)(-5) = 5 > 0\), so \(G(x)\) is positive in \((0, 2)\). For \(x = 3\): \(3(1)(-3) = -9 < 0\), so \(G(x)\) is negative in \((2, 6)\). For \(x = 7\): \(7(5)(1) = 35 > 0\), so \(G(x)\) is positive in \((6, +\infty)\).

Combine the intervals

Combine the intervals where \(G(x)\) is non-negative, including the critical points where \(G(x) = 0\). Thus, \(G(x) \geq 0\) for \([0, 2] \cup [6, +\infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Functions

Polynomial functions play a significant role in mathematics, consisting of terms made up of variables raised to whole number exponents. For example, the polynomial function in our exercise is \(G(x) = x^3 - 8x^2 + 12x\). The highest degree term (in this case, \(x^3\)) determines the overall shape and behavior of the polynomial.

Critical Points

Critical points of a polynomial function are the values of \(x\) where the function either reaches a maximum, minimum, or changes direction. These are usually found by setting the derivative of the function to zero. For our exercise, we set the factored form of \(G(x) = x(x-2)(x-6)\) to zero, leading to \(x = 0\), \(x = 2\), and \(x = 6\) as critical points.

Interval Testing

In interval testing, we investigate the sign of the polynomial function within specific intervals determined by the critical points. This helps us understand where the function is positive or negative. For example, we found that \(G(x)\) < 0 in the intervals \((-\bulletOLegendA, 0)\) and \((2, 6)\), and \(G(x)\) 鈮� 0 in the intervals \([0, 2] \times [6, +\bulletOLegendA]\).

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91影视

For \(G(x)=x^{3}-8 x^{2}+12 x,\) find all \(x\) -values for which \(G(x) \geq 0\).

Short Answer

Step by step solution

Factorize the polynomial

Identify the critical points

Test intervals between critical points

Combine the intervals

Key Concepts

Polynomial Functions

Critical Points

Interval Testing

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