Let f(x) be a polynomial of degree 3 such tha | vedclass

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(B) Let $f(x) = ax^3 + bx^2 + cx + d$. Then $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.Given $f'(x)$ has a critical point at $x = 1$,so $f''(1)

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$3$

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(B) Let $f(x) = ax^3 + bx^2 + cx + d$. Then $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.Given $f'(x)$ has a critical point at $x = 1$,so $f''(1) = 0 \implies 6a(1) + 2b = 0 \implies b = -3a$.Now $f'(x) = 3ax^2 - 6ax + c$. Since $f(x)$ has a critical point at $x = -1$,$f'(-1) = 0 \implies 3a(-1)^2 - 6a(-1) + c = 0 \implies 3a + 6a + c = 0 \implies c = -9a$.So $f(x) = ax^3 - 3ax^2 - 9ax + d$.Using $f(-1) = 10 \implies a(-1)^3 - 3a(-1)^2 - 9a(-1) + d = 10 \implies -a - 3a + 9a + d = 10 \implies 5a + d = 10$.Using $f(1) = -6 \implies a(1)^3 - 3a(1)^2 - 9a(1) + d = -6 \implies a - 3a - 9a + d = -6 \implies -11a + d = -6$.Subtracting the two equations: $(5a + d) - (-11a + d) = 10 - (-6) \implies 16a = 16 \implies a = 1$.Then $b = -3, c = -9$,and $d = 5$. Thus $f(x) = x^3 - 3x^2 - 9x + 5$.$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$.Critical points are $x = -1$ and $x = 3$.$f''(x) = 6x - 6$. At $x = -1$,$f''(-1) = -12 At $x = 3$,$f''(3) = 18 - 6 = 12 > 0$ (local minima).Therefore,$f(x)$ has a local minima at $x = 3$.

Let $f(x)$ be a polynomial of degree $3$ such that $f(-1) = 10$,$f(1) = -6$,$f(x)$ has a critical point at $x = -1$,and $f'(x)$ has a critical point at $x = 1$. Then $f(x)$ has a local minima at $x =$

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