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Everyone. I've been trying to solve this task for an hour now and I got pretty nervous so i had to ask for help. The task is to find the parameter $m$ ($m$ belongs to the set of real numbers) for which the function $y=(m+1)x^2 + (m-5)|x| +m-2$ has the function minimum for $x=1$. I tried to use $x=1$ in the function formula and then use canonical form of quadratic function to get $\alpha =\frac{m-5}{2(m+1)}=-1$ and the for $x>0$, the $m$ is $1$ and for the $x<0$ the $m=-1$. But I am not sure if this solution is correct since when I draw the function it doesn't look as it should.

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    yes, sorry ....2012-03-22

2 Answers 2

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Let $f(x)$ be our function. For positive $x$, we have f'(x)=2(m+1)x+m-5. If we have a local minimum at $x=1$, we must have f'(1)=0. This implies that $2(m+1)+m-5=0$, or equivalently $m=1$. The value of $f(1)$ is then $-3$.

The only thing we need to worry about is what happens for negative $x$. Our curve has obvious symmetry about the $y$-axis. So for $m=1$, an absolute minimum value of $-3$ is also reached when $x=-1$.

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Since you are interested around $x=1$ you can drop the absolute value sign, then take the derivative with respect to $x$ getting y'=2(m+1)x+(m-5) Now you want to find the $m$ that makes this zero at $x=1$, so $0=2(m+1)+m-5$ or $m=1$. You can see the result at this Alpha page