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Home » College Playground » Calculus - Real Analysis » Calculus Computations and Tutorials

2 basic problems about derivative
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Harry Potter
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2 basic problems about derivative

1, Let $f$ is a continuous function in $[a,b]$ and $f$ have derivation on $(a,b)$ such that : $f(a) = f(b)=0$ . Show that with all real number ( $\alpha$ ) we always have $x\in (a,b)$ such that : $\alpha.f(x) + f'(x) = 0$

2, Let $f$ is a continuous function in $[a,b]$ with $a > 0$ and $f$ have derivation on $(a,b)$ . Show that we have $x_{1} \in (a,b)$ such that : $\frac {bf(a) - af(b)}{b - a} = f(x_{1}) - x_{1}f'(x_{1})$

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Last edited by Harry Potter on Sat Oct 31, 2009 1:32 am; edited 1 time in total

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Posted: Fri Oct 30, 2009 8:53 pm

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pco
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Re: 2 basic problems about derivative

Harry Potter wrote:

1, Let $f$ is a continuous function in $[a,b]$ and $f$ have derivation on $(a,b)$ such that : $f(a) = f(b)$ . Show that with all real number ( $\alpha$ ) we always have $x\in (a,b)$ such that : $\alpha.f(x) + f'(x) = 0$

Wrong : choose $f(x)=1$ $\forall x\in[a,b]$ and obviously, for all $\alpha\ne 0$ , we cant find $x\in (a,b)$ such that : $\alpha.f(x) + f'(x) = 0$

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Posted: Fri Oct 30, 2009 11:52 pm

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Harry Potter
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Re: 2 basic problems about derivative

pco wrote:

Harry Potter wrote:

1, Let $f$ is a continuous function in $[a,b]$ and $f$ have derivation on $(a,b)$ such that : $f(a) = f(b)$ . Show that with all real number ( $\alpha$ ) we always have $x\in (a,b)$ such that : $\alpha.f(x) + f'(x) = 0$

Wrong : choose $f(x) = 1$ $\forall x\in[a,b]$ and obviously, for all $\alpha\ne 0$ , we cant find $x\in (a,b)$ such that : $\alpha.f(x) + f'(x) = 0$

i am sorry , i've edited the post , we must add the condition : $f(a)=f(b)=0$ . Shocked

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Posted: Sat Oct 31, 2009 1:34 am

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pco
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Re: 2 basic problems about derivative

Harry Potter wrote:

1, Let $f$ is a continuous function in $[a,b]$ and $f$ have derivation on $(a,b)$ such that : $f(a) = f(b) = 0$ . Show that with all real number ( $\alpha$ ) we always have $x\in (a,b)$ such that : $\alpha.f(x) + f'(x) = 0$

Consider $g(x) = f(x)e^{\alpha x}$ : $g(x)$ is $C_1$ and $g(a) = g(b)$ so $\exists x_1\in(a,b)$ such that $g'(x_1) = 0$ $\iff$ $(f'(x_1) + \alpha f(x_1))e^{\alpha x_1} = 0$

Hence the result.

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Posted: Sat Oct 31, 2009 3:39 am

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