Answer:
Here, the given function,
[tex]f(x) = -x^3 + 3x^2 + x - 3[/tex]
Since, the leading coefficient is negative, and degree is odd,
Thus, the end behaviour of the function is,
[tex]f(x)\rightarrow \infty\text{ as }x\rightarrow -\infty[/tex]
[tex]f(x)\rightarrow -\infty\text{ as }x\rightarrow \infty[/tex]
Therefore, the graph rises to the left and falls to the right.
Now, when f(x) = 0
[tex]-x^3+3x^2+x-3=0[/tex]
[tex]\implies -(x-3)(x-1)(x+1)=0[/tex]
[tex]\implies x=3, 1, -1[/tex]
That is, graph intercepts the x-axis at (3, 0), (1, 0) and (-1, 0).
When x = 0,
[tex]f(x) = - 3[/tex]
That is, graph intersects the y-axis at ( 0, -3),
Also, for 0 > x > -1 , f(x) is decreasing,
For 2.55 > x > 0, f(x) is increasing,
For 3 > x > 2.55, f(x) is decreasing,
Hence, by the above explanation we can plot the graph of the function ( shown below )
