unknown diff. fcn: f"(x)>0, f(2)=10, f(3)=20; find p

[Sendell]

Ok, so I am given that f is a twice-differentiable function with f''(x) > 0 for all real numbers x. Also, f(2) = 10, and f(3) = 20. I am then asked what are possible values for f(4).

I know that the function is always concave up, so f(4) has to be greater than f(3). What else can I assume?

Thanks.

 

[skeeter]

f(4) - f(3) > f(3) - f(2)

 

[pka]

f(4) - f(3) > f(3) - f(2)

OH REALLY?
\(\displaystyle \L
f(x) = \frac{1}{{x^2 }}\)

 

[skeeter]

f(4) - f(3) > f(3) - f(2)

OH REALLY?
\(\displaystyle \L
f(x) = \frac{1}{{x^2 }}\)

yes, oh really.

it was given that f"(x) > 0 and f(3) > f(2) ... f(x) must continue to increase at an increasing rate, otherwise it changes concavity, violating f"(x) > 0 for all x in its domain.

1/x² exhibits the same behavior on the interval where it is increasing, (-infinity, 0).

 

[Sendell]

Haha, yes... Skeeter is right on this one.

Thanks!