Question

describe the end behavior of the graph of $f(x)=x^{3}(x + 3)(-5x + 1)$ using limits. as $x\to-infty,f(x)\to+infty$ as $x\to+infty,f(x)\to-infty$ as $x\to-infty,f(x)\to-infty$ as $x\to+infty,f(x)\to-infty$ as $x\to-infty,f(x)\to-infty$ as $x\to+infty,f(x)\to+infty$ as $x\to-infty,f(x)\to+infty$ as $x\to+infty,f(x)\to+infty$

Explanation:

Step1: Expand the function

First, expand $f(x)=x^{3}(x + 3)(-5x + 1)$.
\[

$$\begin{align*} f(x)&=x^{3}( - 5x^{2}+x-15x + 3)\\ &=x^{3}(-5x^{2}-14x + 3)\\ &=-5x^{5}-14x^{4}+3x^{3} \end{align*}$$

\]

Step2: Find $\lim_{x

ightarrow-\infty}f(x)$
As $x
ightarrow-\infty$, consider the leading - term $-5x^{5}$. When $x
ightarrow-\infty$, $x^{5}
ightarrow-\infty$, and $-5x^{5}
ightarrow+\infty$. Since the leading - term dominates as $x
ightarrow\pm\infty$, $\lim_{x
ightarrow-\infty}(-5x^{5}-14x^{4}+3x^{3})=+\infty$.

Step3: Find $\lim_{x

ightarrow+\infty}f(x)$
As $x
ightarrow+\infty$, the leading - term $-5x^{5}$. When $x
ightarrow+\infty$, $x^{5}
ightarrow+\infty$, and $-5x^{5}
ightarrow-\infty$. So, $\lim_{x
ightarrow+\infty}(-5x^{5}-14x^{4}+3x^{3})=-\infty$.

Answer:

As $x
ightarrow-\infty,f(x)
ightarrow+\infty$
As $x
ightarrow+\infty,f(x)
ightarrow-\infty$