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

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(B) Since $f(x)$ is a polynomial of degree $3$,$f'(x)$ is a quadratic and $f''(x)$ is a linear function.Given $f'(x)$ has a critical point at $x = 1$,

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$3$

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(B) Since $f(x)$ is a polynomial of degree $3$,$f'(x)$ is a quadratic and $f''(x)$ is a linear function.Given $f'(x)$ has a critical point at $x = 1$,we have $f''(1) = 0$. Thus,$f''(x) = k(x - 1)$ for some constant $k$.Integrating $f''(x)$,we get $f'(x) = \frac{k}{2}x^2 - kx + C$.Since $f(x)$ has a critical point at $x = -1$,$f'(-1) = 0$. Substituting $x = -1$: $\frac{k}{2} + k + C = 0 \Rightarrow C = -\frac{3k}{2}$.So,$f'(x) = \frac{k}{2}(x^2 - 2x - 3) = \frac{k}{2}(x - 3)(x + 1)$.Integrating $f'(x)$,we get $f(x) = \frac{k}{2}(\frac{x^3}{3} - x^2 - 3x) + d = \frac{k}{6}x^3 - \frac{k}{2}x^2 - \frac{3k}{2}x + d$.Using $f(1) = -6$: $\frac{k}{6} - \frac{k}{2} - \frac{3k}{2} + d = -6$ $\Rightarrow -\frac{11k}{6} + d = -6$ $\Rightarrow -11k + 6d = -36$ $(i)$.Using $f(-1) = 10$: $-\frac{k}{6} - \frac{k}{2} + \frac{3k}{2} + d = 10$ $\Rightarrow \frac{5k}{6} + d = 10$ $\Rightarrow 5k + 6d = 60$ (ii).Subtracting $(i)$ from (ii): $16k = 96 \Rightarrow k = 6$.Substituting $k = 6$ into (ii): $5(6) + 6d = 60$ $\Rightarrow 6d = 30$ $\Rightarrow d = 5$.Thus,$f(x) = x^3 - 3x^2 - 9x + 5$.The critical points are where $f'(x) = 0$,i.e.,$3(x - 3)(x + 1) = 0$,so $x = 3$ and $x = -1$.$f''(x) = 6x - 6$. At $x = 3$,$f''(3) = 18 - 6 = 12 > 0$,so $f(x)$ 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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