[tex]\begin{gathered} \text{ Given} \\ f\mleft(x\mright)=\frac{x^4}{4}-\frac{2x^3}{3}-\frac{x^2}{2}+2x \end{gathered}[/tex]
Graphing the function f(x) over the interval [-2.5, 2.3], we have the following
x-intercepts
Judging on the graph, the x-intercept of the function f(x) are the following
[tex]x=-1.619\text{ and }x=0[/tex]
critical points
Finding the first derivative and setting it to zero and solve for x to get the critical points in the interval [-2.5,2.3] we have
[tex]\begin{gathered} f\mleft(x\mright)=\frac{x^4}{4}-\frac{2x^3}{3}-\frac{x^2}{2}+2x \\ f^{\prime}\left(x\right)=x^3-2x^2-x+2 \\ \\ x^3-2x^2-x+2=0 \\ \left(x−2\right)\left(x+1\right)\left(x−1\right)=0 \end{gathered}[/tex]
Therefore, the critical points are x = 2, x = 1, x = -1.
relative minimums
Basing on the graph our relative minimum is at x = -1, and x = 2
asymptotes
Since the given function is a polynomial function, the function has no asymptotes
critical numbers
same as critical points with x = 2, x = 1, and x = -1.
relative maximum
observing the graph, the relative maximum of the function is at x = 1.
inflection points
get the second derivative of the function and set it to zero
[tex]\begin{gathered} f^{\prime}\left(x\right)=x^3-2x^2-x+2 \\ f^{\prime}^{\prime}\left(x\right)=3x^2-4x-1 \\ 3x^2-4x-1=0 \end{gathered}[/tex]
Using the quadratic formula we have the values at
[tex]x=\frac{2}{3}\pm\sqrt{\frac{7}{3}}[/tex]
end behavior
Since the function is a polynomial with a degree of 4, and a positive leading coefficient, the end behavior of the function is increasing in both approaching to -2.5, and 2.3.
