Let f(x) be a cubic polynomial with f(1) = -1 | vedclass

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(C) Let $f(x) = ax^3 + bx^2 + cx + d$. Then $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.Since $f'(x)$ has a local minima at $x = -1$,$f''( -1) =

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$22$

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(C) Let $f(x) = ax^3 + bx^2 + cx + d$. Then $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.Since $f'(x)$ has a local minima at $x = -1$,$f''( -1) = 0$,which gives $6a(-1) + 2b = 0$,so $b = 3a$.Thus,$f'(x) = 3ax^2 + 6ax + c = 3a(x^2 + 2x) + c = 3a(x+1)^2 + (c - 3a)$.Since $f(x)$ has a local minima at $x = 1$,$f'(1) = 0$,so $3a(1+1)^2 + (c - 3a) = 0$,which gives $12a + c - 3a = 0$,so $c = -9a$.Now,$f'(x) = 3ax^2 + 6ax - 9a = 3a(x^2 + 2x - 3) = 3a(x+3)(x-1)$.Integrating $f'(x)$,we get $f(x) = a(x^3 + 3x^2 - 9x) + k$.Using $f(1) = -10$: $a(1 + 3 - 9) + k = -10 \Rightarrow -5a + k = -10$.Using $f(-1) = 6$: $a(-1 + 3 + 9) + k = 6 \Rightarrow 11a + k = 6$.Subtracting the equations: $(11a + k) - (-5a + k) = 6 - (-10) \Rightarrow 16a = 16 \Rightarrow a = 1$.Then $k = 6 - 11(1) = -5$.So,$f(x) = x^3 + 3x^2 - 9x - 5$.Finally,$f(3) = (3)^3 + 3(3)^2 - 9(3) - 5 = 27 + 27 - 27 - 5 = 22$.

Let $f(x)$ be a cubic polynomial with $f(1) = -10$,$f(-1) = 6$,and it has a local minima at $x = 1$. Also,$f'(x)$ has a local minima at $x = -1$. Then $f(3)$ is equal to:

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