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### Course: AP®︎/College Calculus AB>Unit 5

Lesson 4: Using the first derivative test to find relative (local) extrema

# Analyzing mistakes when finding extrema (example 2)

Analyzing the work of someone who tried to find extrema of a function, to see whether they made mistakes.

## Want to join the conversation?

• I thought critical points cannot be 0 in the original function. If you were to plug in 1 or -1 into the original function it would be 0, so would -1 and 1 be voided critical points? Making the assumption right at the end?
• No, this is not true. Critical values points that make the original function undefined, not zero, are voided because they're not in the domain of the function. Clearly, critical points where the function is 0 are in the domain of the function and so would not be voided. (Think of it this way: there would be no reason to void the critical point (minimum) at 0 for the function x^2, just because this function's value happens to be 0 there.)

Note that the original function f(x) = (x^2 - 1)^(2/3), at 1 and -1, is 0 instead of undefined, since the exponent 2/3 is positive. So these critical points 1 and -1 would not be voided.
• Do we always need to find all critical points? What about functions like
x^3/(x+sin(x)) ?
• You don't need to find all the critical points if the x value that makes the derivative 0 also makes the original function zero. For example 1/x.

Also, you don't need to find critical points of regular trig functions since they're repeating.
(1 vote)
• Isn't the f' also undefined for x € [0, 1] since the square root can't be negative?
• For Step 3, in the third column, first row, the equation reads "`f'(3) = -2 < 0`". Since this is the evalution of the test x-value, x= -3, shouldn't this read "`f'(-3) = -2 < 0`"?