The discussion centers on proving that if the derivative of a function $$f(x)$$ is zero, then the function itself must be constant. Participants reference the Mean Value Theorem, which states that if a function is continuous and differentiable on an interval, then there exists a point where the derivative equals the average rate of change. If $$f'(x) = 0$$ for all $$x$$, it follows that $$f(a) = f(b)$$ for any two points $$a$$ and $$b$$ in the interval, confirming that $$f(x)$$ is constant.
PREREQUISITESMathematics students, educators, and anyone interested in understanding the foundational concepts of calculus, particularly in relation to derivatives and constant functions.
suvadip said:If $$f(x)$$ is constant then I can show that $$f'(x)=0$$.
But if $$f'(x)=0$$, then how to show that $$f(x)$$ is constant ?
The mean value theorem: if f is continuous and differentiable on [a, b] then there exist c in [a, b] such that f'(c)= \frac{f(b)- f(a)}{b- a}. If f'(x)= 0 for all x, then for any a and b, \frac{f(b)- f(a)}{b- a}= 0[/tex] from which f(a)= f(b).suvadip said:If $$f(x)$$ is constant then I can show that $$f'(x)=0$$.
But if $$f'(x)=0$$, then how to show that $$f(x)$$ is constant ?