MHB Derivative of Constant: Proving f(x) is Constant

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If a function f(x) is constant, it can be proven that its derivative f'(x) equals zero. Conversely, if f'(x) equals zero, it can be shown that f(x) must also be constant by applying the Mean Value Theorem. The theorem states that if f is continuous and differentiable on an interval, there exists a point where the derivative equals the average rate of change, which leads to the conclusion that f(a) equals f(b) for any two points a and b. Additionally, using differentials reinforces the idea that a zero derivative implies no change in the function's value. Thus, both directions of the argument confirm that a function with a zero derivative is indeed constant.
Suvadip
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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 ?
 
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Re: derivative of constant

What does the definition of the derivative tell you?

edit: Sorry, I misread...have you tried integrating with respect to the independent variable?
 
Last edited:
Re: derivative of constant

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 ?

Another approach would be to use differentials:
$$\Delta y=f'(x)\, \Delta x = 0 \cdot \Delta x \, ...$$
 
Re: derivative of constant

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).
 
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