# 1. Let f(x) = ax3 + 6x2 +bx + 4. Determine the constants a and b so that f has a relative minimum at x = -1 and a relative maximum at x = 2. Show all working!

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`f(x)=ax^3+6x^2+bx+4`

`f'(x)=3ax^2+12x+b`

`f''(x)=6ax+12`

Since f(x) has minimum value at x=-1. Therefore

`f''(-1)=-6a+12>0`

`2>a`

a<2 (i)

And f(x) has maximum value at x=2, therefore

`f''(2)=12a+12<0`

`a<-1` (ii)

Thus from (i) and (ii)

a<-1

Since x=-1 and x=2 are relative minima and relative maxima, therefore

`f'(-1)=0`

`f'(2)=0`

Thus

3a-12+b=0 (iii)

12a+24+b=0 (iv)

Subtract (iii) from (iv),

9a+36=0

a=-4

Substitute a=-4 in (iii), we have b=24.

Thus we have a=-4 and b=24.