properties of functions

• Oct 30th 2006, 10:15 PM
gracy
properties of functions
state true or false
1. If f has absolute minimum value at c then f '(c)=0
2. If f is contineous on(a,b) then f attains an absolute maximum value f(c) and absolute minimum value f(d) at some numbers c and d in (a,b)
3.If f is differentiable and f(-1)=f(1) then there is a number c such that |c|<0 and f ' (c)=0.
4.If f ' (x)=0 for 1<x<6 then f is decreasing on (1,6)
5.If f " (x) =0 then (2,f(2) is an inflection point of curve y=f(x).
6.If f ' (x) = g ' (x) for all 0<x<1 , then f(x) =g(x) for 0<x<1
• Oct 31st 2006, 12:41 AM
Glaysher
Quote:

Originally Posted by gracy
state true or false
1. If f has absolute minimum value at c then f '(c)=0
2. If f is contineous on(a,b) then f attains an absolute maximum value f(c) and absolute minimum value f(d) at some numbers c and d in (a,b)
3.If f is differentiable and f(-1)=f(1) then there is a number c such that |c|<0 and f ' (c)=0.
4.If f ' (x)=0 for 1<x<6 then f is decreasing on (1,6)
5.If f " (x) =0 then (2,f(2) is an inflection point of curve y=f(x).
6.If f ' (x) = g ' (x) for all 0<x<1 , then f(x) =g(x) for 0<x<1

Not 100% sure on these but I'm sure someone else will check

1. False, f may only be defined for an interval and c may be one of the ends of the interval

2. True

3. False |c|< 0 not possible (unless you are using a different definition of |c|

4. False. f ' (x) = 0 for 1 < x < 6 means f(x) is constant on (1, 6)

5. False (though I think this one may be mistyped and could be true)

6.False eg f(x) = 2x + 6, g(x) = 2x - 5
f '(x) = 2 = g'(x)
But f(0.5) = 7 and g(0.5) = -4
• Oct 31st 2006, 01:41 AM
TD!

1) It's possible that f'(c) isn't defined at the minimum x = c. Example: f(x)=|x| with c = 0.
2) That would depend on how you defined "absolute max/min", if it has to be strict, consider a constant function.
3) I think you mean |c| < 1. Since f is diff, it is also continuous, thus it is true by Rolle's theorem.
4) See previous reply; or true if your definition of 'decreasing' doesn't have a strict inequality.
5) You probably mean f"(x) = 0 for at x = 2. This isn't sufficient for an inflection point, f"(x) has to change sign there.
6) See previous reply; functions with equal derivatives are only equal up to an arbitrary constant.
• Oct 31st 2006, 05:33 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
2) That would depend on how you defined "absolute max/min", if it has to be strict, consider a constant function.

All the texts I have gazed upon define a relative maximum/minimum as point $c$ such that for some open interval containing $c$ the function is defined and $f(x)\leq f(c)$, $f(x)\geq f(c)$ respectively. Meaning that a line both is a relative maximum and minimum point. The reason why such a defintion was chosen is because when we prove the Extreme Value Theorem we do not have to trouble ourselfs with functions that are not strict, thus it becomes easier in the proof.

Quote:

4) See previous reply; or true if your definition of 'decreasing' doesn't have a strict inequality.
I think that function is constant on that open interval.
• Oct 31st 2006, 10:16 AM
TD!
@2: I assume you mean "each point on a (straight) line is a relative min/max" and not the line itself. Given this definition, not using strict inequalities, that's true.
@4: The line would indeed be constant, but notice the analogy with the definition above. If you take increasing/decreasing as non-strict inequalities, a line is both decreasing and increasing. Again a matter of definition.