F(x)= ax+3 ; $0 \leq x < \frac{1}{2}$ $0≤x<1/2$ f(x)= $\sqrt{\frac{2 x - 1}{x + 15}}$ $sqrt((2x-1)/(x+15))$ ; $\frac{1}{2} \leq x \leq 1$ $1/2≤x≤1$ f(x)= $(\sqrt{x + 3} - 2)$ $(sqrt(x+3)-2)$ /(x-1) ; $x > 1$ $x>1$ Find the value of a that makes f(x) continuous in $\frac{1}{2}$ $1/2$ ?

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F(x)= ax+3 ; $0 \leq x < \frac{1}{2}$ $0≤x<1/2$ f(x)= $\sqrt{\frac{2 x - 1}{x + 15}}$ $sqrt((2x-1)/(x+15))$ ; $\frac{1}{2} \leq x \leq 1$ $1/2≤x≤1$ f(x)= $(\sqrt{x + 3} - 2)$ $(sqrt(x+3)-2)$ /(x-1) ; $x > 1$ $x>1$ Find the value of a that makes f(x) continuous in $\frac{1}{2}$ $1/2$ ?

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Answer 1 · 2017-09-13T11:50:07

$a = - 6$ $a=color(red)(-6)$

Explanation:

Since $f (x) = \sqrt{\frac{2 x - 1}{x + 15}}$ $f(x)=sqrt((2x-1)/(x+15))$ for $\frac{1}{2} \leq x < 1$ $1/2 <= x < 1$

at $x = \frac{1}{2}$ $x=1/2$
$XXX f (\frac{1}{2}) = \sqrt{\frac{2 \cdot \frac{1}{2} - 1}{\frac{1}{2} + 15}} = 0$ $color(white)("XXX")f(1/2)=sqrt((2 * 1/2 -1)/(1/2+15))=color(magenta)0$

If $f (x) = a x + 3$ $f(x)=color(blue)ax+3$ for $x < \frac{1}{2}$ $x < 1/2$

to be continuous
as $x \to \frac{1}{2}$ $x rarr 1/2$ then $a x + 3 \to 0$ $color(blue)ax+3 rarr color(magenta)0$

that is, as $x \to \frac{1}{2}$ $xrarr 1/2$
$XXX a \cdot \frac{1}{2} + 3 = 0$ $color(white)("XXX")color(blue)a * 1/2 +3 = color(magenta)0$

$XXX a + 6 = 0$ $color(white)("XXX")color(blue)a+6 =color(magenta)0$

$XXX a = - 6$ $color(white)("XXX")color(blue)a=color(red)(-6)$

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F(x)= ax+3 ; $0 \leq x < \frac{1}{2}$ $0≤x<1/2$ f(x)= $\sqrt{\frac{2 x - 1}{x + 15}}$ $sqrt((2x-1)/(x+15))$ ; $\frac{1}{2} \leq x \leq 1$ $1/2≤x≤1$ f(x)= $(\sqrt{x + 3} - 2)$ $(sqrt(x+3)-2)$ /(x-1) ; $x > 1$ $x>1$ Find the value of a that makes f(x) continuous in $\frac{1}{2}$ $1/2$ ?

1 Answer

Explanation:

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F(x)= ax+3 ; 0≤x<120≤x<1/2 f(x)= √2x−1x+15sqrt((2x-1)/(x+15)) ; 12≤x≤11/2≤x≤1 f(x)= (√x+3−2)(sqrt(x+3)-2)/(x-1) ; x>1x>1 Find the value of a that makes f(x) continuous in 121/2 ?

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F(x)= ax+3 ; $0 \leq x < \frac{1}{2}$ $0≤x<1/2$ f(x)= $\sqrt{\frac{2 x - 1}{x + 15}}$ $sqrt((2x-1)/(x+15))$ ; $\frac{1}{2} \leq x \leq 1$ $1/2≤x≤1$ f(x)= $(\sqrt{x + 3} - 2)$ $(sqrt(x+3)-2)$ /(x-1) ; $x > 1$ $x>1$ Find the value of a that makes f(x) continuous in $\frac{1}{2}$ $1/2$ ?