Consider j(x)= 12x^3/5- 27x/5. Describe and provide an explanation for the end behavior of j(x).

Answer 1

To analyze the end behavior of the function $j(x)=\frac{12}{5}{x}^3-\frac{27}{5}x$, we focus on the leading term, since it will dominate the behavior of the function as $x$ approaches positive or negative infinity.

The leading term of the function is $\frac{12}{5}{x}^3$. We consider the following:

1. Degree of the polynomial: The polynomial $j(x)$ is of degree 3 (since the highest exponent of $x$ is 3).
2. Leading coefficient: The leading coefficient is $\frac{12}{5}$, which is positive.

End Behavior Analysis:

• As $x\to \mathrm{\infty }$ (positive infinity):

  • The term $\frac{12}{5}{x}^3$ dominates. Since it is a cubic term with a positive coefficient, $j(x)$ will approach $+\mathrm{\infty }$.
  • Therefore, $j(x)\to +\mathrm{\infty }$ as $x\to +\mathrm{\infty }$.

• As $x\to -\mathrm{\infty }$ (negative infinity):

  • Again, the leading term $\frac{12}{5}{x}^3$ dominates. In this case, as $x$ becomes more negative, $x^3$ also becomes more negative, thus $j(x)$ will approach $-\mathrm{\infty }$.
  • Therefore, $j(x)\to -\mathrm{\infty }$ as $x\to -\mathrm{\infty }$.

Conclusion:

In summary, the end behavior of the function $j(x)$ can be stated as follows:

• As $x\to +\mathrm{\infty }$, $j(x)\to +\mathrm{\infty }$.
• As $x\to -\mathrm{\infty }$, $j(x)\to -\mathrm{\infty }$.

This behavior is typical for an odd-degree polynomial with a positive leading coefficient.