Let \[$ G \$\] Be The Function Given By:$\[ g(x) = \begin{cases} 3x^2 - 5x + 1 & \text{if } X \leq 0 \\ \frac{1}{x(x+1)} & \text{if } X \ \textgreater \ 0 \end{cases} \\]Which Of The Following Statements Is True?A. \[$ G

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Introduction

In mathematics, a piecewise function is a function that is defined by multiple sub-functions, each applied to a specific interval of the domain. The function g(x)g(x) given by the piecewise definition is a classic example of such a function. In this article, we will analyze the function g(x)g(x) and discuss its properties, including its domain, range, and behavior at different intervals.

The Piecewise Function g(x)g(x)

The function g(x)g(x) is defined as:

g(x)={3x2−5x+1if x≤01x(x+1)if x>0g(x) = \begin{cases} 3x^2 - 5x + 1 & \text{if } x \leq 0 \\ \frac{1}{x(x+1)} & \text{if } x > 0 \end{cases}

Domain and Range

To determine the domain and range of the function g(x)g(x), we need to analyze the two sub-functions separately.

Domain of g(x)g(x)

The domain of the first sub-function, 3x2−5x+13x^2 - 5x + 1, is all real numbers, since it is a polynomial function. However, the domain of the second sub-function, 1x(x+1)\frac{1}{x(x+1)}, is restricted to x>0x > 0, since the function is undefined when x=0x = 0 or x=−1x = -1. Therefore, the domain of g(x)g(x) is x≤0x \leq 0 or x>0x > 0, excluding x=0x = 0 and x=−1x = -1.

Range of g(x)g(x)

The range of the first sub-function, 3x2−5x+13x^2 - 5x + 1, is all real numbers, since it is a polynomial function. However, the range of the second sub-function, 1x(x+1)\frac{1}{x(x+1)}, is restricted to positive real numbers, since the function is always positive. Therefore, the range of g(x)g(x) is all real numbers, excluding non-positive real numbers.

Behavior at Different Intervals

To analyze the behavior of the function g(x)g(x) at different intervals, we need to examine the two sub-functions separately.

Behavior at x≤0x \leq 0

At x≤0x \leq 0, the function g(x)g(x) is defined by the sub-function 3x2−5x+13x^2 - 5x + 1. This is a quadratic function, which has a parabolic shape. The function has a minimum value at x=−56x = -\frac{5}{6}, and the minimum value is −112-\frac{1}{12}.

Behavior at x>0x > 0

At x>0x > 0, the function g(x)g(x) is defined by the sub-function 1x(x+1)\frac{1}{x(x+1)}. This is a rational function, which has a vertical asymptote at x=0x = 0 and x=−1x = -1. The function has a horizontal asymptote at y=0y = 0.

Which of the Following Statements is True?

Now that we have analyzed the function g(x)g(x), we can examine the given statements and determine which one is true.

A. g(x)g(x) is a continuous function.

B. g(x)g(x) is a differentiable function.

C. g(x)g(x) has a maximum value at x=0x = 0.

D. g(x)g(x) has a minimum value at x=−56x = -\frac{5}{6}.

E. g(x)g(x) has a vertical asymptote at x=0x = 0.

F. g(x)g(x) has a horizontal asymptote at y=0y = 0.

Conclusion

In conclusion, the function g(x)g(x) is a piecewise function that is defined by two sub-functions. The function has a domain of x≤0x \leq 0 or x>0x > 0, excluding x=0x = 0 and x=−1x = -1. The function has a range of all real numbers, excluding non-positive real numbers. The function has a minimum value at x=−56x = -\frac{5}{6}, and the minimum value is −112-\frac{1}{12}. The function has a horizontal asymptote at y=0y = 0.

Answer

The correct answer is D. g(x)g(x) has a minimum value at x=−56x = -\frac{5}{6}.

References

Q: What is the domain of the function g(x)g(x)?

A: The domain of the function g(x)g(x) is x≤0x \leq 0 or x>0x > 0, excluding x=0x = 0 and x=−1x = -1.

Q: What is the range of the function g(x)g(x)?

A: The range of the function g(x)g(x) is all real numbers, excluding non-positive real numbers.

Q: What is the behavior of the function g(x)g(x) at x≤0x \leq 0?

A: At x≤0x \leq 0, the function g(x)g(x) is defined by the sub-function 3x2−5x+13x^2 - 5x + 1. This is a quadratic function, which has a parabolic shape. The function has a minimum value at x=−56x = -\frac{5}{6}, and the minimum value is −112-\frac{1}{12}.

Q: What is the behavior of the function g(x)g(x) at x>0x > 0?

A: At x>0x > 0, the function g(x)g(x) is defined by the sub-function 1x(x+1)\frac{1}{x(x+1)}. This is a rational function, which has a vertical asymptote at x=0x = 0 and x=−1x = -1. The function has a horizontal asymptote at y=0y = 0.

Q: Is the function g(x)g(x) continuous?

A: Yes, the function g(x)g(x) is continuous, since it is defined by two sub-functions that are continuous at the point of intersection.

Q: Is the function g(x)g(x) differentiable?

A: No, the function g(x)g(x) is not differentiable at the point of intersection, since the two sub-functions have different derivatives at that point.

Q: Does the function g(x)g(x) have a maximum value at x=0x = 0?

A: No, the function g(x)g(x) does not have a maximum value at x=0x = 0, since the function is undefined at that point.

Q: Does the function g(x)g(x) have a minimum value at x=−56x = -\frac{5}{6}?

A: Yes, the function g(x)g(x) has a minimum value at x=−56x = -\frac{5}{6}, and the minimum value is −112-\frac{1}{12}.

Q: Does the function g(x)g(x) have a vertical asymptote at x=0x = 0?

A: Yes, the function g(x)g(x) has a vertical asymptote at x=0x = 0, since the sub-function 1x(x+1)\frac{1}{x(x+1)} is undefined at that point.

Q: Does the function g(x)g(x) have a horizontal asymptote at y=0y = 0?

A: Yes, the function g(x)g(x) has a horizontal asymptote at y=0y = 0, since the sub-function 1x(x+1)\frac{1}{x(x+1)} approaches 0 as xx approaches infinity.

Conclusion

In conclusion, the function g(x)g(x) is a piecewise function that is defined by two sub-functions. The function has a domain of x≤0x \leq 0 or x>0x > 0, excluding x=0x = 0 and x=−1x = -1. The function has a range of all real numbers, excluding non-positive real numbers. The function has a minimum value at x=−56x = -\frac{5}{6}, and the minimum value is −112-\frac{1}{12}. The function has a horizontal asymptote at y=0y = 0.

References