calculus help..pic below
For local minimum, f'(x) will go from negative to positive, becoming 0 at the minimum point. For local maximum, f'(x) will go from positive to negative, becoming 0 at the maximum point. Here f'(x) = 0 at x = -1 and x = 0. So x = -1 and 0 are potential maximum/minimum points. Take x = -1: f'(x) is negative for x < -1 and f'(x) is positive on the open interval (-1,0) So f'(x) goes from negative to positive. Therefore, x = -1 is a minimum. Take x = 0: f'(x) is positive on the open interval (-1,0), becomes 0 at x = 0, and f'(x) is positive on the open interval (0, infinity). f'(x) does not change sign at x = 0 So f'(x) goes from negative to positive. Therefore, no local extrema at x = 0.
oh okay that makes sense thank you:)
The last line above should be: "So f'(x) goes from POSITIVE to positive" You are welcome.
any idea how to do #3?
I am having problems solving this one: the function f whose derivative is given by f'(x)=5x^3-15x+7 has a local max at x= A) -1.930 B)-1 C)0.511 D)1 E)1.419
put each one of the choices above into f'(x) and see which one makes it zero. If more than one choice makes it zero, find f''(x) and put the x value and see if it is negative to make sure it is a local maximum.
okay i will try that
doe sit have to equal exactly 0? cause when i plugged in -1.930 i got 4.715x10^-3
There will be rounding off errors. Are you allowed to use graphing calculator?
yes
Plot f'(x) = 5x^3-15x+7 and see where all it goes to zero. Note: Be careful. Don't confuse the maximum / minimum of f'(x) in the graph with the maximum and minimum of f(x). We are interested in where f'(x) is zero to locate the critical points. Then do the second derivative test to see which is max.
A C and E are close to zero
correct. Now find f''(x) and test each point.
it has to equal a negative answer?
for max, f''(x) < 0
0.511 :)
yes!!
yay!:)
You did good. :)
i think i know how to do number 3 now :)
I did not notice #3. But if you got it then that is good.
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