Relative Minimum, value of $a$ For what value of a does the function $f(x) = x^2 + ax$ have a relative minimum at $x = 1$?
I don't know where to begin on this problem.
 A: $$f(x) = x^2 + ax$$
To find critical points (which are values of $x$ at which there may be local minima or maxima), we first determine whether there are any values of $x$ for which the function is not defined (here, f(x) is defined everywhere), and then: 


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*Find the first derivative $f'(x)$



 $$f'(x) = 2x + a$$



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*Set $f'(x) = 0$, and solve for $x$.



 $$f'(x) = 0 \iff 2x + a = 0 \iff x = -\frac a2.$$ To determine what value of $a$ would make $x = 1$ a possible extrema/critical point, we set $-\dfrac a2 = x = 1\iff a = -2$.



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*We test each solution $x$ that solves $f'(x) = 0$ to see whether/where it is a local minima or a local maxima. 


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*We can do this using the first derivative to determine where the function is increasing and/or decreasing. 




 Note that we found earlier that the only way $x = 1$ can possibly be a critical point is if $a = -2$. So $x = 1$ is a critical point if and only if $f(x) = x^2 - 2x$ and  $f'(x) = 2x - 2$. Note, that when $x\lt 1$, $f'(x)<0$ (f(x) is decreasing). And when $x\gt 1, \;f'(x) >0 \implies f(x)$ is increasing. Hence, when $a = -2$, $x = 1$ is indeed  a local maximum. 

A: To a minimum corresponds a first derivative equal to zero and a second derivative positive. The first derivative of you function is 2 x + a; for which value of "a" and x=1 is this derivative equal to zero ? Remember to check that this point corresponds to a minimum.
A: let us  derivative given function,we get  $f'(x)=2*x+a  $  ,so    $f'(x)=0--> 2*x+a=0 $ 
or  
$a=-2*x$
putting we get $f(x)=x^2-2*x^2$ ,which is equal $f(x)=-x^2$  so that $f(1)=-1$
by the we  have to use second derivative test to check that it is really minimum, second derivative of given function is  $2$,because it is positive,then $a=-2*x$ is really minimum,if it would be negative,then we should have maximum point
