To find the leading term degree and coefficient we need to expand the function:
[tex]\begin{gathered} f(x)=(x+1)(3x-1)(4x+1)^3 \\ =(3x^2-x+3x-1)((4x)^3+3(4x)^2(1)+3(4x)(1)^2+1^3) \\ =(3x^2+2x-1)(64x^3+3(16x^2)+12x+1) \\ =(3x^2+2x-1)(64x^3+48x^2+12x+1) \\ =192x^5+144x^{4^{}}+36x^3+3x^2+128x^4+96x^3+24x^2+2x-64x^3-48x^2-12x-1 \\ =192x^5+272x^4+68x^3-21x^2-10x-1 \end{gathered}[/tex]From this we conclude that the polynomial is of degree 5, that is the leading term has degree 5 and its coefficient is 192.
The graph of the function is:
From this we notice that:
The function increases in the intervals:
[tex]\begin{gathered} (-\infty,-0.833) \\ \text{and} \\ (0.2,\infty) \end{gathered}[/tex]The function decreases in the interval:
[tex](-0.833,02)[/tex]Furthermore as the variable tends to minus infinity the function also tends to minus infinity; whereas as the variable tends to infinity the function tends to inifnity, that is:
[tex]\begin{gathered} \lim _{x\to-\infty}f(x)=-\infty \\ \lim _{x\to\infty}f(x)=\infty \end{gathered}[/tex]
