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
+118608 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.
im guessing its gonna be 2 b/c maxmium increase is 6, and 2 is not in its range, is this right
+118608 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.
+118608 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. :)
+128408 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
