Math problem: I need a function f(x) such that the derivative > 0 but the rate of change is decreasing.?

Question

I feel like this is not possible.

Answer 1

rate of change is decreasing means f'' < 0.
f' is already the rate of change.

one example of this: f(x) = ln(x). x > 0

f'(x) = 1/x ... this is positive for positive x

f''(x) = -1/x^2 ... always negative.

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Another example... where the domain is the set of all real numbers...
h(x) = 1 - e^(-x)
h'(x) = e^(-x) > 0
h''(x) = -e^(-x) < 0

Generally, all curves that are increasing, yet , concave down...

Answer 2

Plenty of examples, that just means that they want something where the derivative is strictly positive, but converges to 0 as n approaches infinity.

Example 1:
f(x)=ln(x), x>0
=>f'(x)=1/x >0
and f'(1)=1, f'(2)=1/2 so the rate of change is decreasing

Example 2
or f(x)=x/(x+1)
f'(x)=1/(x+1)^2
f'(1) = 1/4, f'(2)=1/9, so again the rate of change is decreasing


graph either of those functions to further see what they mean.

Answer 3

a curve function such as 3x^2+4x-4 has D<0 and a>0. therefore this function always lie above the x-axis. and the derivative of this function is always >0 since 6x+4 >0. but the rate of change is decreasing until the min. point of the curve. hope I have solved your problem.

Answer 4

You probably can't have this for all x in your domain, but you can have it for some x in your domain. Consider y= -x^2 as x=>0 from the negative side.

Answer 5

try f(x)=sqrt(x) the derivative will always be positive but as x gets bigger, the value of the derivative decreases

Answer 6

f(x) = -e^(-x)

f(x) = ln (x),

for example.......