In a problem like, Find the critical point of $$f(x)= \frac{5x}{x-3}$$ What happens to the $5x$ in the numerator? I see that my book has the answer as 3. in order to get 3 I see that I can set $x-3=0$ and add 3 to both sides correct $x-3+3=0+3$ Therefore $x=3$. Is this the correct way to look at this problem?
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$\begingroup$Usually you would define a "critical point" as the zero-points of the first derivative of your function. Since this functions first derivative has no zero-point, the critical point you search for is probably the point where your function is not defined. This means for your example to find the zero-points of the denominator, because it is "not allowed" to divide by 0. Your posted solution does exactly this and hence is the correct way to look at this problem.
$\endgroup$ 1 $\begingroup$In order to find the critical points, you have the derivative first.
Your function is $$f(x)= \frac{5x}{x-3}$$
Assuming you know the quotient rule, the derivative will then become $$f'(x) = -\frac{15}{(x-3)^2}$$
Critical points are defined as points where either $f'(x) = 0$ or $f'(x)$ is undefined.
Looking at the derivative, we see that when $$x=3$$
$f'(x)$ is undefined
You misinterpreted the answer. When it said x =3 was a critical point, you assumed that it came from $f(x)$. In your context, x =3 would be a vertical asymptotic for $f(x).$
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