If the functions f(x) = frac{x^3}{3} + 2bx + | vedclass

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(D) For a function to have an extreme point,its derivative must be zero at that point.$f'(x) = x^2 + ax + 2b$$g'(x) = x^2 + 2bx + a$Let $x_0$ be the c

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$6$

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(D) For a function to have an extreme point,its derivative must be zero at that point.$f'(x) = x^2 + ax + 2b$$g'(x) = x^2 + 2bx + a$Let $x_0$ be the common extreme point. Then $f'(x_0) = 0$ and $g'(x_0) = 0$.$x_0^2 + ax_0 + 2b = 0$  ---$(1)$$x_0^2 + 2bx_0 + a = 0$  ---$(2)$Subtracting equation $(2)$ from $(1)$:$(a - 2b)x_0 + (2b - a) = 0$$(a - 2b)x_0 - (a - 2b) = 0$$(a - 2b)(x_0 - 1) = 0$Since $a 
eq 2b$,we must have $x_0 - 1 = 0$,which implies $x_0 = 1$.Substituting $x_0 = 1$ into equation $(1)$:$1^2 + a(1) + 2b = 0$$1 + a + 2b = 0$$a + 2b = -1$We need to find the value of $a + 2b + 7$.Substituting $a + 2b = -1$ into the expression:$-1 + 7 = 6$.

If the functions $f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$ and $g(x) = \frac{x^3}{3} + ax + bx^2$,where $a \neq 2b$,have a common extreme point,then $a + 2b + 7$ is equal to

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