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Question

If f(x)=a(x2-1)(ax+b),a0, has local maximum and minimum at x=1 and x=2 respectively, then f'(x) is:


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Solution

Step 1: Solve for f'(x)

Given: f(x)=a(x2-1)(ax+b),a0 has a local maximum and minimum at x1 and x2.

f(x)=a(x2-1)(ax+b)ddxf(x)=ddxa(x2-1)(ax+b)f(x)'=addx(x2-1)(ax+b)

applying the product rule of differentiation

f'(x)=a(ax+b)ddx(x2-1)+(x2-1)ddx(ax+b)f'(x)=a(ax+b)·2x+(x2-1)af'(x)=a2ax2+2bx+ax2-af'(x)=a3ax2+2bx-a

Step 2: Solve for f'(1)

putting x=1

f'(1)=a3a×(1)2+2b(1)-af'(1)=a3a+2b-af'(1)=2a(a+b)

Step 3: Solve for f'(2)

putting x=2

f'(2)=a3a×(2)2+2b(2)-af'(2)=a12a+4b-af'(2)=a(11a-4b)

Hence,f'(x) for x=1,2 is 2a(a+b) and a(11a-4b).


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