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Hello Pawan:

Could you solve this problem for me algebraically:

g(x)= 3f(x+1)-2 and also if you could list the transformations of f(x)= x^3-2x+x to obtain g(x)

Thank you.

Angelo

Angelo,

f(x) = x^3-2x+x

Hence f(x+1) = (x+1)^3 - 2(x+1) + (x+1) ....... (Replacing x with x+1 in f(x))

f(x+1) = [x^3 + 1^3 + 3*x^2*1+ 3*x*1^2] - 2x-2 +x+1 ...... [(a + b)^3 = a^3 + b^3 + 3ab(a + b)]

= x^3 + 1 + 3x^2 + 3x

= x^3 + 3x^2 + 3x +1 ..........(After rearranging above equation)

Thus we get, f(x+1) = x^3 + 3x^2 + 3x +1 ...................................(1)

Now, g(x)= 3f(x+1)-2

Putting value of f(x+1) from equation (1) above in g(x), we get,

g(x) = 3(x^3 + 3x^2 + 3x +1) - 2

g(x) = 3x^3 + 9x^2 + 9x + 3 - 2

g(x) = 3x^3 + 9x^2 + 9x + 1 ................................(2)

I hope that this is clear. Please let me know if you need further clarification.

All the best!

Regards,

Pawan

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