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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 1)

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

## Want to join the conversation?

• Isn't this a trick question? Does x^3-6*x^2+12x actually have any relative extrema?
• No, it doesn't. It's derivative is 3x²-12x+12, which factors as 3(x-2)², which is 0 at 2 and positive elsewhere. Since extrema require the derivative to switch between positive and negative, this function has no extrema.
• so any point where the derivative is 0 is called a critical point right??
(1 vote)
• For a continuous function, anywhere in the domain of the function where the derivative is either 0 or undefined is called a critical point.
• This isn't really related to the topic, but if we had an equation like
12x^2 + 4x - 4
Can we factor out a 4x from the terms even though only two of them have it?
• No, I recommend reviewing algebra. What you can do is factor out the 4.
12x^2 + 4x - 4 = 4(3x^2 + x + 1)
• we could have assumed the equation doesnt have an extreme by just looking at the derivative of the equations also, right?

3x^2-> would be an upward facing parabola. meaning it would not have a max point only a min point.
(1 vote)
• I'm a little confused. So when you go from a peak at 3, then drop back down to 0, then go back up to 3, that's not a curve? What happens with the line?Isn't the graph you showed a value of -3? Any help would be appreciated!
(1 vote)
• The key is in the fact that we are working with derivatives. Even though the derivative decreases from 3 to 0 and then back to 3, it never becomes less than zero. Since taking the derivative is just the rate of change/ slope at those points, the slope is always positive over this interval, except for at zero where the tangent line is just horizontal (dy/dx = 0).

However, if we were taking the actual value of the function and got points (1,3) and (2,0), you are right, the derivate would be negative at any point where 1<x<2.
(1 vote)