So I need help with Session 28 from Part A of 2. Applications of Differentiation...?

So in page 3 of the notes, I understand that f'(0+) and x=0+ is a critical point...but in the lecture professor Jerison mentions that it's also the horizontal slope? Why is that?

Question

So I need help with Session 28 from Part A of 2. Applications of Differentiation...?

So in page 3 of the notes, I understand that f'(0+) and x=0+ is a critical point...but in the lecture professor Jerison mentions that it's also the horizontal slope? Why is that?

kutulu · Answer

All critical points have horizontal slope. A critical point is one where the derivative $$f'(x)$$ is zero. Since the value of the derivative is the slope of the tangent line, anywhere that $$f'(x) = 0$$ means the line must be horizontal. 

When we try to compute the value of the derivative at x = 0 we get infinities, so we have to visualize what happens as x gets very very small: ln(x) approaches negative infinity. So: 1/ln(x) approaches 0, so we have:$$f'(0+) = (1/\ln0+) + (1/(\ln0+)^2) = 0+0 = 0$$

This just means that we can picture the x < 1 case as a curve with it's maximum at x=0, except that we have to remove the left "half" because it cannot exist.

anonymous · Answer

oh okay that makes sense. Thank you!!