Let f(x) = begin{cases} frac{2^x + 2^{3-x} - | vedclass

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(C) To check for continuity at $x = 2$,we evaluate the limits from both sides.For $x > 2$,$f(x) = \frac{2^x + 8 \cdot 2^{-x} - 6}{2^{-x/2} - 2 \cdot 2

Vedclass · Accepted Answer

$f(2^-) 
eq f(2^+) \Rightarrow f$ is discontinuous

Vedclass · Answer

(C) To check for continuity at $x = 2$,we evaluate the limits from both sides.For $x > 2$,$f(x) = \frac{2^x + 8 \cdot 2^{-x} - 6}{2^{-x/2} - 2 \cdot 2^{-x}}$. Let $u = 2^{-x/2}$,then $2^x = u^{-2}$ and $2^{-x} = u^2$. As $x 	o 2^+$,$u 	o 2^{-1} = 1/2$. The expression becomes $\frac{u^{-2} + 8u^2 - 6}{u - 2u^2} = \frac{1 + 8u^4 - 6u^2}{u^2(u - 2u^2)} = \frac{(2u^2 - 1)(4u^2 - 1)}{u^2(u(1 - 2u))} = \frac{-(2u^2 - 1)(2u - 1)(2u + 1)}{u^3(2u - 1)} = \frac{-(2u^2 - 1)(2u + 1)}{u^3}$. Substituting $u = 1/2$,we get $\frac{-(2(1/4) - 1)(1 + 1)}{1/8} = \frac{-( -1/2)(2)}{1/8} = 8$. Thus,$f(2^+) = 8$.For $x Since $f(2^+) 
eq f(2^-)$,the function has a jump discontinuity at $x = 2$.

Let $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.

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