Given range and first derivative values, find maximum pt

Dec 2014
7
0
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Hello. I was thinking about this kind of problem...

range [0,4]

0 <= f'(x) <=3

what is the maximum point?

so, I assume x=3 when f'(x)=4 (assuming some sort of f(x) curving upwards with the tangent slopes increasing from 0). Not sure how to get y from there.

maybe tanget line...y=mx+b...y=(4)(3)+b...y=12+b

but then what is the y-intercept, b? two unknowns.

been struggling with this without a function. Thank you for any help.
 

skeeter

Math Team
Jul 2011
3,360
1,850
Texas
Hello. I was thinking about this kind of problem...

range [0,4]

0 <= f'(x) <=3

what is the maximum point?

so, I assume x=3 when f'(x)=4 ...
how can $f'(x)=4$ if the derivative values are $0 \le f'(x) \le 3$ ?

If the range of a function is $[0,4]$, then the maximum value of the function is $y=4$ ... also, the given interval of values for $f'$ indicates that $f(x)$ never decreases throughout its domain.
 

mathman

Forum Staff
May 2007
6,932
774
Hello. I was thinking about this kind of problem...

range [0,4]

0 <= f'(x) <=3

what is the maximum point?

so, I assume x=3 when f'(x)=4 (assuming some sort of f(x) curving upwards with the tangent slopes increasing from 0). Not sure how to get y from there.

maybe tanget line...y=mx+b...y=(4)(3)+b...y=12+b

but then what is the y-intercept, b? two unknowns.

been struggling with this without a function. Thank you for any help.
Careful with terminology. Range means y values, domain means x values. I'll assume you meant domain [0,4]. Then the maximum is at f(4), but with no further information, that is all you can say.

If you really meant range [0,4], then 4 is the maximum value.