For ` - 1 le x le 2` , we have
`" " f(x) = 3x ^(2) + 12 x - 1 `
`rArr " " f' (x) = 6x + 12 gt 0 , AA - 1 le x le 2 `
Again, function is an algebraic polynomial, therefore it is continuous at `x in (-1, 2) and ( 2, 3)`.
For continuity at ` x = 2 `,
`" " underset ( x to 2^(-)) (lim) f(x) = underset (x to 2^(-)) (lim) ( 3x ^(2) + 12 x - 1)`
`" " = underset ( h to 0) (lim) [ 3 (2 - h) ^(2) + 12 ( 2 - h) -1]`
`" " = underset ( h to 0 ) (lim) [ 3 ( 4 + h ^(2) - 4h ) + 24 -12 h - 1]`
`" " = underset (h to 0 ) ( lim) ( 1 2 + 3h^(2) - 12 h + 24 - 12h - 1 ) `
`" " = underset ( h to 0 ) ( lim ) ( 3h ^(2) - 24 h + 35)= 35`
and ` underset ( x to 2^(+)) (lim) f(x) = underset ( x to 2 ^(+) ) (lim) ( 37 - x)`
`" " = underset (h to 0) ( lim) [ 37 - ( 2 + h ) ] = 35`
and `f (2) = 3*2^(2) + 12* 2 - 1 = 12 + 24 - 1 = 35 `
Therefore, LHL = RHL `= f (2) to` function is continuous at ` x = 2 rArr ` function is continuous in ` - 1 le x le3`.
Now, `R f' (2) = underset ( x to 2^(+)) (lim) (f(x) - f(2))/( x - 2)`
` " " = underset (h to 0 ) ( lim) ( f(2 + h ) - f (2))/( h )`
`" " = underset( h to 0 )(lim) ( 37 - (2 + h)- ( 3 xx 2^(2) + 12 xx 2 - 1) ) /( h )`
` " " = underset (h to 0 ) ( lim) (-h)/(h) = -1 `
and `Lf' (2) = underset ( x to 2^(-)) ( f(x) - f(2))/( x - 2) = underset ( h to 0) ( lim) ( f(2 - h ) f(2))/( - h ) `
` " " = underset ( h to 0 ) ( lim ) ( [ 3 ( 2 - h ) ^(2) + 12 ( 2- h ) - 1] - ( 3xx 2^(2) + 12 xx 2 - 1)])/( -h )`
`= underset ( h to 0 ) ( lim) ([ 12 + 3h ^(2) - 12 h + 24 - 12 h - 1 ] - 35)/( -h )`
`= underset (h to 0) ( lim)( 3h ^(2) - 24 h + 35 - 35) /( - h)`
` = underset( h to 0) ( lim) ( 3h - 24)/(-1) = 24 `
Since, `R f'(2) ne L f' (2) , f ' (2) ` does not exist.
Again, `f(x)` is an increasing in ` [ - 1, 2]` and is decreasing in `(2, 3)`, it shows that `f (x)` has a maximum value at ` x = 2`.
Therefore, options (a), (b), (c), (d) are all correct.
