If f x -x2-4x-3- x- 2 x-4- x- 2 - g x -x-3- x- 3 x2-2x-2- x- 3 - Determine f-g-1

The critical points are 2, 3 and so we redefine the functions for x lt; 2, 2≤ x lt; 3, x gt; 3. #160; f ( x )={x^2 4x+3, amp;x lt; 2 ...

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Byju's Answer

Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

If f x =x2-...

Question

If $f (x) = {\begin{matrix} x^{2} - 4 x + 3, & x < 2 x - 4, & x \geq 2 \end{matrix}$ $g (x) = {\begin{matrix} x - 3, & x < 3 x^{2} + 2 x + 2, & x \geq 3 \end{matrix}$ Determine $\frac{f}{g} \cdot (1)$

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Solution

The critical points are 2, 3 and so we redefine the functions for $x < 2, 2 \leq x < 3, x > 3.$
$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} - 4 x + 3, & x < 2 x - 4, & 2 \leq x < 3 x - 4, & x \geq 3 \end{matrix}$

$g (x) ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x - 3, & x < 2 x - 3, & 2 \leq x < 3 x^{2} + 2 x + 2, & x > 3 \end{matrix}$
$∴ f + g = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} - 3 x, & x < 2 2 x - 7, & 2 \leq x < 3 x^{2} + 2 x - 2, & x \geq 3 \end{matrix}$

$\frac{f}{g} = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{2} - 4 x + 3}{x - 3} = x - 1, & x < 2 \frac{x - 4}{x - 3}, & 2 \leq x < 3 \frac{x - 4}{x^{2} + 3 x + 2,} & x \geq 3 \end{matrix}$
So, when $x = 1, x \in (2, - \infty)$
$⟹ \frac{f}{g} (1) = 1 - 1 = 0$

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If f(x)={x2−4x+3,x<2x−4,x≥2 g(x)={x−3,x<3x2+2x+2,x≥3 Determine fg⋅(1)

If $f (x) = {\begin{matrix} x^{2} - 4 x + 3, & x < 2 x - 4, & x \geq 2 \end{matrix}$ $g (x) = {\begin{matrix} x - 3, & x < 3 x^{2} + 2 x + 2, & x \geq 3 \end{matrix}$ Determine $\frac{f}{g} \cdot (1)$