Analyzing a Continuous Decreasing Function: Critical Point at (4,2)

name_ask17
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Homework Statement


Multiple Choice If f is a continuous, decreasing function on
[0, 10] with a critical point at (4, 2), which of the following statements
must be false? E
(A) f (10) is an absolute minimum of f on [0, 10].
(B) f (4) is neither a relative maximum nor a relative minimum.
(C) f ' (4) does not exist.
(D) f ' (4) = 0
(E) f ' (4) < 0



Homework Equations



Ok. It looks to me like C and E are both false, based on the mere fact that D is correct, making this question have 2 answers.
Can someone please explain to me why one of them should be incorrect?


The Attempt at a Solution



Thanks in advance!
 
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How did you define critical point??
 
critical point was where f '(x)= 0, so i said f '(4)=0
 
name_ask17 said:
critical point was where f '(x)= 0, so i said f '(4)=0

Under that definition, it seems indeed true that C and E are false. However, I would doublecheck that definition if I were you.
 
Last edited:
micromass said:
Under that definition, it seems indeed true that C and E are true. However, I would doublecheck that definition if I were you.

No. A function can be *strictly decreasing* and yet have a critical point. For example, f(x) = -x^3 is strictly decreasing but has f&#039;(0) = 0. (It is strictly decreasing because for any x_1 &lt; x_2 we have f(x_1) &gt; f(x_2).)

RGV
 
Yeah, critical points aren't just where the derivative is zero...
 
Ray Vickson said:
No. A function can be *strictly decreasing* and yet have a critical point. For example, f(x) = -x^3 is strictly decreasing but has f&#039;(0) = 0. (It is strictly decreasing because for any x_1 &lt; x_2 we have f(x_1) &gt; f(x_2).)

RGV

Yes, how does that contradict what I said??
 
micromass said:
Yes, how does that contradict what I said??

Sorry: it doesn't; I did not read the questions A--E carefully enough.

RGV
 
Ok, so the answer is E then.
Thanks!
 
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