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

[name_ask17]

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!

 

[micromass]

How did you define critical point??

 

[name_ask17]

critical point was where f '(x)= 0, so i said f '(4)=0

 

[micromass]

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.

 

[Ray Vickson]

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 $f'(0) = 0$. (It is strictly decreasing because for any $x_1 < x_2$ we have $f(x_1) > f(x_2).$)

RGV

 

[Mark44]

Yeah, critical points aren't just where the derivative is zero...

 

[micromass]

No. A function can be *strictly decreasing* and yet have a critical point. For example, $f(x) = -x^3$ is strictly decreasing but has $f'(0) = 0$. (It is strictly decreasing because for any $x_1 < x_2$ we have $f(x_1) > f(x_2).$)

RGV

Yes, how does that contradict what I said??

 

[Ray Vickson]

Yes, how does that contradict what I said??

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

RGV

 

[name_ask17]

Ok, so the answer is E then.
Thanks!