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let f be a differentiable function such that f(0)=-5 and f'(x) 3 for all x, but which value is not possible for f(2)?

 

-10

-5

0

1

2

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Guest Feb 6, 2016

Best Answer 

 #2
avatar+91001 
+10

let f be a differentiable function such that f(0)=-5 and f'(x)= 3 for all x, but which value is not possible for f(2)?

 

what you must remember is that f'(x) is the gradient of the tangent to the curve.  If this gradient is any constant then the curve MUST be a line.

SO

This is a line with a gradient ot 3 and a y intercept of -5

the equation has to be

f(x)=3x-5

 

f(2)=2*3-5 = 1

 

No other value is possible.

Melody  Feb 6, 2016
edited by Melody  Feb 6, 2016
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5+0 Answers

 #1
avatar
+5

im guessing its gonna be 2 b/c maxmium increase is 6, and 2 is not in its range, is this right 

Guest Feb 6, 2016
 #2
avatar+91001 
+10
Best Answer

let f be a differentiable function such that f(0)=-5 and f'(x)= 3 for all x, but which value is not possible for f(2)?

 

what you must remember is that f'(x) is the gradient of the tangent to the curve.  If this gradient is any constant then the curve MUST be a line.

SO

This is a line with a gradient ot 3 and a y intercept of -5

the equation has to be

f(x)=3x-5

 

f(2)=2*3-5 = 1

 

No other value is possible.

Melody  Feb 6, 2016
edited by Melody  Feb 6, 2016
 #3
avatar
+5

im sorry its suppose to be f'(x) lesser or equal to 3

Guest Feb 6, 2016
 #4
avatar+91001 
+5

ok then, if the gradient was equal to 3 then f(2)=1

 

if the gradient is less than 3 then f(2) must be less than 1 (because it is not so steep anywhere)

 

so f(2) cannot be more than 1.    Only one of those points is more than 1.   :)

Melody  Feb 6, 2016
 #5
avatar+78618 
+10

Thanks, Melody.......!!!  ......here's another way to see this...

 

This must be a linear equation if the derivative is just some integer at all x values...so....

 

At  0, f(x)  = -5

 

So....at f(2).......x  has changed by 2  units

 

Thus, using the formula for slope :

 

 [ f(2) - (-5) ] / [ 2  - 0]  <=  3     simplify

 

[ f(2)  + 5]  / 2  <= 3     multiply both sides by 2

 

f(2) + 5 <= 6      subtract 5 from both sides

 

f(2)  <=  1

 

Thus f(2) cannot  = 2   because this is greater than 1

 

 

cool cool cool

CPhill  Feb 6, 2016

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