Suppose that #f'(x) = 2x# for all #x#. What is #f(2)# if #f(1)=0#? What if #f(-2) = 3#?

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Answer 1 · 2016-12-28T22:23:31

Knowing the derivative, we should recognize that its definite integral generates an arbitrary constant:

#int 2xdx = x^2 + C#

Given the initial condition #f(1) = 0#, we have:

#(1)^2 + C = 0#

#C = -1#

so that

#f(x) = x^2 - 1#

and

#color(blue)(f(2)) = (2)^2 - 1 = color(blue)(3)#

Or, given a different initial condition #f(-2) = 3#, we have:

#(-2)^2 + C = 3#

#C = -1#

so that again,

#f(x) = x^2 - 1#

and

#color(blue)(f(2)) = (2)^2 - 1 = color(blue)(3)#

It just means that both initial conditions are based on the same antiderivative (including its y-intercept).

What if #f(0) = -1#? How would you solve it then? :-)