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For part d)
1 if f"(k)>0 , f has a local minimum at x=k
2 if f"(k)<0 , f has a local maximum at x=k
3 if f(x)=ax^3+bx^2+cx+d then
4 f'(x)=3ax^2+2bx+c and
5 f"(x)=6ax+2b
6 Substituting x=-2 into line 5 and setting the result to greater than 0 will give
7 us constraints for a and b
8 -12a+2b>0
9 b>6a
10 Substituting x=4 into line 5 and setting the result to smaller than 0 will give
11us constraints for a and b
12 24a+2b<0
13 b<-12a
14 From this we know that -12a>6a , implying that a<0
15 a=-1 fits this constraint, so we can use this
16 6a<b<-12a
17 -6<b<12
18 b=3 fits this constraint, so we can use this
19 The derivative of f(x) is equal to 0 at x=-2 and x=4
20 3ax^2+2bx+c=0
21 Substituting values of a and b and x=-2 into line 20 , we get
22 12(-1)-4(3)+c=0 giving that
23 c=24
24 We now have an equation that has a local minimum at x=-2 and a local maximum at x=4
25 -(x^3)+3(x^2)+24x