# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for f(x) = 2x + ln x?

Mar 22, 2017

This function increases on the interval: (-infinity,-1/2)u(0,infinity) and decreases on the interval: ($- \frac{1}{2}$,0). The relationship ative maxima occur at x=$- \frac{1}{2}$ and there are no relative minima.

#### Explanation:

To figure this out we must first take the derivative of the function which is $\frac{1}{x} + 2$. Wherever the derivative of the function is positive, the function is increasing, and wherever it is negative, the function is decreasing. If you know what the parent function, $\frac{1}{x}$ looks like, then this should be easy to determine. However this is translated up two units from the parent function so although it usually would not intersect the x-axis it will here so we need to determine where it will. To do this we set $\frac{1}{x} + 2 = 0$. When we solve for x we get x=$- \frac{1}{2}$. This means that the derivative will be positive from -infinity to $- \frac{1}{2}$ and from 0 to infinity (that second part is found via examining the parent function). This also where the function will increase.

Secondly, the derivative is negative from the x-intercept to 0. So the function decreases from $- \frac{1}{2}$ to 0.

Relative maxima occur at x-intercepts of the derivative where it goes from positive to negative (because this is where the function goes from increasing to decreasing) and this only occurs at $x = - \frac{1}{2}$ so that is where the relative maxima is.

Relative minima occur at x-intercepts of the derivative where it goes from negative to positive (because this is where the function goes from decreasing to increasing). This never occurs due to the vertical asymptote so there are no relative minima for this function. Hope I helped! Sorry if this was confusing, I was a little rushed!