ok to use the first derivative test you test between the interval
Now plug in values and test, if you go from - to + you have a minimum and if you go from + to - you have a maximum.
On that question I found the stationary points, but I don't know how to do the first derivative test. I have the formula but my function give me headache to work with can someone help me.
Find any stationary points of the function g(x) defined in part (a)(ii), and use the First derivative Test to classify each stationary point as a local maximum or a local minimum of g(x).
square root 10+x^2
This equation has solutions x= 5 and x= 2. Thus the stationery points of f are at x= 5 and x= 2.
To classify the stationary point at x= 5, we choose test points xl= -1 and xr= 1, say. Then we have
f’(xl) = f’(-1)=
Ok so the power went out due to hurricane IKE so im going to try to answer this the best i can on my cellphone lol. Plug in x = 3 and x = 6 into the 1st derivative and see what values it gives you. Go ahead and try x = 0 too to test all the intervals.
ok i figured out what you mean by xl and xr. using your method we need to test the critical points 2 and 5. so to test for 2 xl would be 0 and and xr would b 3. to test 5 xl would b 3 and xr would be 7. you will notice that when you test for 2 xl will be positive and xr will be negative. when u test for 5 xl will be negative and xr will be positive. so 2 a max and 5 a min. hopes this helps. ill write and explain it more clearly when i get power back. it just a tad difficult to type this on a cellphone
Ok time to finish this up lol
you found the derivative
Now set the derivative to 0 to find critical points
so the critical points are
Now using the 1st derivative test we test for our intervals
ok now we try a x value for each of our intervals
so lets try 0 for x < 2
g'(0) = .04518
try 3 for 2 < x < 5
g'(3) = -.0053
try 6 for 5 < x
g'(6) = .00432
Ok from this you can see that the XL side of 2 is positive and the XR side is negative making x = 2 a max. You can also see that Xl side of 5 is negative and the XR side is positive making x = 5 a min. You have to remember to stay within the intervals when picking your XL and XR. For example for the critical point 5, I could not chose XL to be 0 because it is not in the interval of 2 < x < 5
Hope this makes sense