If f x - - -1 - kx - -1 - kx-x for - 1 - x - 0- 2x2 - 3x - 2 for 0 - x - 1- - is continuous at x-0 then k-

x→ 0 ^ + Lim #160; #160; f( x ) = 2( 0 ) #160; ^ 2 +3( 0 ) 2= 2 x→ 0 ^ Lim #160; #160; f( x ) = x→ 0 ^ Lim ( √( 1+kx ) √( ...

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Theorems for Continuity

If f x = [ ...

Question

If $f (x) = {\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} f o r - 1 \leq x < 0 2 x^{2} + 3 x - 2 f o r 0 \leq x < 1 \end{matrix}$
is continuous at $x = 0$ then $k =$

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f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - x}}{x} & f o r - l \leq x < 0 2 x^{2} + 3 x - 2 & f o r 0 \leq x \leq 1 \end{matrix}

x = 0

k

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} & f o r & - 1 \leq x < 0 2 x^{2} + 3 x - 2 & f o r & 0 \leq x < 1 \end{matrix}

x = 0

=

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & - 1 \leq x < 0 \end{matrix} \begin{matrix} 2 x^{2} + 3 x - 2, & 0 \leq x \leq 1 \end{matrix} \end{matrix}

x = 0

k

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} & i f - 1 \leq x < 0 \frac{2 x + 1}{x - 1} & i f 0 \leq x < 1 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

x = 0

f (x)

x = 0

f (x) = \frac{8^{x} - 2^{x}}{k^{x} - 1}

x \neq 0

= 2,

x = 0

k

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