Let f(x) be a polynomial of degree 3 such that
Question:

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 $f^{\prime}(x)$ has a critical point at $x=1$. Then $f(x)$ has a local minima at $x=$

Solution:

Let $f(x)=a x^{3}+b x^{2}+c x+d$

$f(-1)=10$ and $f(1)=-6$

$-a+b-c+d=10$……..(1)

$a+b+c+d=-6$……..(2)

Solving equations (i) and (ii), we get

$a=\frac{1}{4}, d=\frac{35}{4}$

$b=\frac{-3}{4}, c=-\frac{9}{4}$

$\Rightarrow f(x)=a\left(x^{3}-3 x^{2}-9 x\right)+d$

$f^{\prime}(x)=\frac{3}{4}\left(x^{2}-2 x-3\right)=0$

$\Rightarrow x=3,-1$

Local minima exist at $x=3$

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