Sketch the graph of the following functions `y=f(x)` and find the number of real roots of the corresponding equation `f(x)=0`.
`(i) f(x)=2x^(3)-9x^(2)+12x-(9//2) " " (ii) f(x)=2x^(3)-9x^(2)+12x-3`

Explanation

Let `f(x) = 2x^(3) - 9x^(2) + 12x - (9//2)`. Then
`f(x) = 6x^(2) - 18x +12`
`= 6(x^(2)-3x+2) = 6(x- 1)(x-2)`
Now `f'(x) = 0 rArr` x = 1 and x = 2
Also `f(1) = 2 - 9 + 12 - (9//2) gt 0`
and `f(2) = 16 - 36 + 24 - (9//2) lt 0`
Hence, the graphs of the function `y= f(x) is as shown in the figure.
(##CEN_ALG_C02_SLV_026_S01.png" width="80%">
As shown in the figure, the graph cuts the x-axis at three distinct points
Hence, equation`f(x) = 0` has three distinct roots.
(ii) For `2x^(3) - 9x^(2) + 12x - 3 = 0,`
`f(x) = 2x ^(3) - 9x + 12x - 3`
`f'(x)= 0`
`rArr 6x^(2) - 18x + 12 = 0`
or 6 (x - 1) (x - 2) = 0
`rArr` x = 1 and x = 2
Also ` f(1) = 2 - 9 + 12 - 3 = 2`
and `f(2) = 16 - 36 + 24 - 3 = 1`
Hence, the graph of y = f(x) is as shown in the figure.

Thus , from the graph, we can see that f(x) = 0 has only one real root, though y = f(x) has two turning points .