If x=-1 and x=2 are extreme points of [f( x ) = , | x |+ x^2 + x] then

Question

If $x=-1$ and $x=2$ are extreme points of $$f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$$ then

Accepted Answer

Let $$f\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x$$

Differentiating both sides,

$$f'\left( x \right) = {\alpha  \over x} + 2\beta x + 1$$

Since $x=-1$ and $x=2$ are extreme points therefore

$$f'\left( x \right) = 0$$ at these points.

Put $x =  - 1$ and $x = 2$ in $$f'\left( x \right),$$

we get $$ - \alpha  - 2\beta  + 1 = 0$$

$$ \Rightarrow \alpha  + 2\beta  = 1\,\,...\left( i \right)$$

$${\alpha  \over 2} + 4\beta  + 1 = 0$$

$$ \Rightarrow \alpha  + 8\beta  =  - 2\,\,...\left( {ii} \right)$$

On solving $(i)$ and $(ii)$, we get

$$6\beta  =  - 3 \Rightarrow \beta  =  - {1 \over 2}$$

$$\therefore$$ $$\,\,\,\,\alpha  = 2$$

Solutions · 1