1. ## first derivative test

Hi Everyone,

I'm new in the forum.

I would like to know if my exercise is correct, often I make mistake in the calculation. If there is a way I can check my work.

Here is the question:

*Using the strategy to apply the First Derivative Test, classify the
left-hand stationary point:
stationary points:
X= -3 and x= 1
.*

Step 1: the left stationery point of f is at x= -3;

Step 2: To classify the stationery point at x= -3, we choose test points xl= -4 and xr= -2, say. Then we have
f ’(xl)= f ’(-4) = 3(-4)(-1)(-5) = -60 < 0
f ’(xr)= f ’(-2) = 3(-2)(1)(-3) = 18 > 0

so f has a local minimum at x = -2, by the first derivative test.

I hope there is no mistake in that

2. Originally Posted by valerie-petit
Hi Everyone,

I'm new in the forum.

I would like to know if my exercise is correct, often I make mistake in the calculation. If there is a way I can check my work.

Here is the question:

*Using the strategy to apply the First Derivative Test, classify the
left-hand stationary point:
stationary points:
X= -3 and x= 1
.*

Step 1: the left stationery point of f is at x= -3;

Step 2: To classify the stationery point at x= -3, we choose test points xl= -4 and xr= -2, say. Then we have
f ’(xl)= f ’(-4) = 3(-4)(-1)(-5) = -60 < 0
f ’(xr)= f ’(-2) = 3(-2)(1)(-3) = 18 > 0

so f has a local minimum at x = -2, by the first derivative test. Mr F says: Should that be x = -3 .....? Otherwise it all looks OK. Keeping in mind of course that you have not given the rule for the function so the values themselves can't actually be checked ......

I hope there is no mistake in that
..