Average rate of change

Question

Show that the average rate of the change of the function f(X) = x^2 is greater than the average rate of change of the function f(X) =X on every interval [a,b] In [1,∞]. Is this true if we allow [a,b] to be in [0, ∞).

Answer 1

For f(x) = x², the average rate of change of f on [a,b] is                              

 (f(b)-f(a))/(b-a) = (b²-a²)/(b-a) = (b-a)(b+a)/(b-a) = b+a

 

For f(x) = x, the average rate of change of f on [a,b] is (b-a)/(b-a) = 1

 

b+a > 1 when a,b ≥ 1

 

So, the average rate of change of f(x) = x² on [a,b] is greater than the average rate of change of f(x)=x on [a,b] for every interval [a,b] in [1,∞].

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The above is not necessarily true if we allow a and b to be between 0 and 1.  

 

For example, if a = 1/8 and b = 3/8, then the average rate of change of f(x) = x² on [a,b] is b + a = 1/2, but the average rate of change of   f(x) = x on [a,b] is 1.