need some help on finding the critical value and x intercept of this equation:
y(x) = ax-xln(x)
thanksness
the x-intercept is the $\displaystyle x_0$ such that $\displaystyle y(x_0)=0$. so...
$\displaystyle 0=ax-x\ln x$
$\displaystyle \rightarrow ax=x\ln x$
$\displaystyle \rightarrow a = \ln x$
$\displaystyle \rightarrow x=e^{a}$
so the x-intercept of this function is $\displaystyle (e^{a},0)$.
What's the definition of a critical value? I'm not familiar with that term.
A critical value would be a value where the first derivative of the function is 0 or where the function does not exist.
So the function does not exist at x = 0, so this is a critical point.
$\displaystyle y^{\prime}(x) = a - ln(x) - 1$
So when this is 0 we have a critical point:
$\displaystyle a - ln(x) - 1 = 0$
$\displaystyle ln(x) = a - 1$
$\displaystyle x = e^{a - 1}$
So this is your other critical point.
-Dan