Question #264657

The function f: [ 2,4] to R , defined by f(x)= 3/x is uniformly continuous on its domain.


True or false with full explanation

1
Expert's answer
2021-11-18T06:54:20-0500

Define f(x)=3xx[2,4].f(x)=\frac{3}{x} \forall x \in[2,4].

Then f is continuous and bounded function on [2,4].

For any δ>0\delta>0 , we can find mNm \in \mathbb{N} such that 1n<δ\frac{1}{n}<\delta for every nmn \geq m

Let x1=3m,x2=32mx_{1}=\frac{3}{m}, x_{2}=\frac{3}{2 m} , so that x1,x2[2,4]x_{1}, x_{2} \in[2,4]

Consider x2x1=32m3m=32m>δ2>δ\left|x_{2}-x_{1}\right|=\left|\frac{3}{2 m}-\frac{3}{m}\right|=\frac{3}{2 m}>\frac{\delta}{2}>\delta

and f(x2)f(x1)=2m3m3=m3\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|=|\frac{2 m}{3}-\frac{m}{3}|=\frac{m}{3}

Therefore if we choose ε>0\varepsilon>0 such that ε>m\varepsilon>m

then we get x2x1>δ\left|x_{2}-x_{1}\right|>\delta and f(x2)f(x1)<ε\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\varepsilon

Hence f is uniformly continuous on [2,4].


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