Question

2. a. Prove that f(x) = V22 - 13 is uniformly continuous on the interval (7,0). b. Prove or disprove: f(x) = V x2 - 13 is not uniformly continuous on the interval ( 13, 7). c. Prove or disprove: If a > 13, f(x) = 32-13 is uniformly continuous on the interval (a,

Answer 1

A function f(x) is said to be uniformly continuous on a set X if for every e > 0, 38 >0 such that Vx. ve X. x - y = f(x) = f( u se

.

.

(a) PROOF:

$\\*\text{Let }\epsilon>0. \\*\\*\text{Now, }\forall x,y\in [7,\infty), \\*|f(x)-f(y)| \\*\\*=|\sqrt{x^2-13}-\sqrt{y^2-13}| \\*\\*< |\sqrt{x^2}-\sqrt{y^2}|\quad\quad[\because x^2-13<x^2,\quad y^2-13<y^2] \\*\\*= |x-y|\quad\quad[x\in[7,\infty)\implies x>0\implies \sqrt{x^2}=x] \\*\\*\\*\therefore |f(x)-f(y)|<|x-y|\quad\forall x,y\in [7,\infty) \\*\\*\implies |f(x)-f(y)|<|x-y|< \epsilon\text{ whenever }|x-y|< \epsilon \\*\\*\text{Let }\delta=\epsilon. \\*\\*\boxed{\text{Hence, }\forall x,y\in [7,\infty),\quad |x-y|< \delta =\epsilon\implies |f(x)-f(y)|< \epsilon} \\*\\*\text{Thus, }f(x)\text{ is uniformly continuous on the interval }[7,\infty). \\*\text{Hence proved.}$

.

.

(b) We DISPROVE the statement as follows:

$\\*\text{Let }\epsilon>0. \\*\\*\text{Now, }\forall x,y\in (\sqrt{13},7], \\*|f(x)-f(y)| \\*\\*=|\sqrt{x^2-13}-\sqrt{y^2-13}| \\*\\*< |\sqrt{x^2}-\sqrt{y^2}|\quad\quad[\because x^2-13<x^2,\quad y^2-13<y^2] \\*\\*= |x-y|\quad\quad[x\in(\sqrt{13},7]\implies x>0\implies \sqrt{x^2}=x] \\*\\*\\*\therefore |f(x)-f(y)|<|x-y|\quad\forall x,y\in(\sqrt{13},7] \\*\\*\implies |f(x)-f(y)|<|x-y|< \epsilon\text{ whenever }|x-y|< \epsilon \\*\\*\text{Let }\delta=\epsilon. \\*\\*\boxed{\text{Hence, }\forall x,y\in (\sqrt{13},7],\quad |x-y|< \delta =\epsilon\implies |f(x)-f(y)|< \epsilon} \\*\\*\text{Thus, }f(x)\text{ is uniformly continuous on the interval }(\sqrt{13},7]. \\*\text{Hence the statement that f(x) is not uniformly continuous on } (\sqrt{13},7]\text{ is disproved.}$

.

.

(c) We PROVE the statement as follows:

$\\*\text{Let }\epsilon>0. \\*\\*\text{Now, }\forall x,y\in [a,\infty), \\*|f(x)-f(y)| \\*\\*=|\sqrt{x^2-13}-\sqrt{y^2-13}| \\*\\*< |\sqrt{x^2}-\sqrt{y^2}|\quad\quad[\because x^2-13<x^2,\quad y^2-13<y^2] \\*\\*= |x-y|\quad\quad[x\in[a,\infty)\implies x>0\implies \sqrt{x^2}=x] \\*\\*\\*\therefore |f(x)-f(y)|<|x-y|\quad\forall x,y\in [a,\infty) \\*\\*\implies |f(x)-f(y)|<|x-y|< \epsilon\text{ whenever }|x-y|< \epsilon \\*\\*\text{Let }\delta=\epsilon. \\*\\*\boxed{\text{Hence, }\forall x,y\in [a,\infty),\quad |x-y|< \delta =\epsilon\implies |f(x)-f(y)|< \epsilon} \\*\\*\text{Thus, }f(x)\text{ is uniformly continuous on the interval }[a,\infty). \\*\text{Hence proved.}$​​​​​​​​​​​​​​