Given the functions $f (x) = x^{2} - 4$ $f(x)=x^2 -4$ and $g (x) = 2 x - 3$ $g(x)= 2x -3$ , how do you find the positive interval of unity length, in which these two functions intersect?

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Given the functions $f (x) = x^{2} - 4$ $f(x)=x^2 -4$ and $g (x) = 2 x - 3$ $g(x)= 2x -3$ , how do you find the positive interval of unity length, in which these two functions intersect?

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Answer 1 · 2018-07-02T21:08:48

Two given functions $f (x) = x^{2} - 4$ $f(x)=x^2-4$ & $g (x) = 2 x - 3$ $g(x)=2x-3$ will intersect each other where $f (x) = g (x)$ $f(x)=g(x)$

$x^{2} - 4 = 2 x - 3$ $x^2-4=2x-3$

$x^{2} - 2 x - 1 = 0$ $x^2-2x-1=0$

$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (1) (- 1)}}{2 (1)}$ $x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)}$

$x = 1 \pm \sqrt{2}$ $x=1\pm\sqrt2$

$g (1 + \sqrt{2}) = 2 (1 + \sqrt{2}) - 3 = 2 \sqrt{2} - 1$ $g(1+\sqrt2)=2(1+\sqrt2)-3=2\sqrt2-1$

$g (1 - \sqrt{2}) = 2 (1 - \sqrt{2}) - 3 = - 2 \sqrt{2} - 1$ $g(1-\sqrt2)=2(1-\sqrt2)-3=-2\sqrt2-1$

The functions intersect each other at two points $(\sqrt{2} + 1, 2 \sqrt{2} - 1)$ $(\sqrt2+1, 2\sqrt2-1)$ & $(- \sqrt{2} + 1, - 2 \sqrt{2} - 1)$ $(-\sqrt2+1, -2\sqrt2-1)$

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Given the functions $f (x) = x^{2} - 4$ $f(x)=x^2 -4$ and $g (x) = 2 x - 3$ $g(x)= 2x -3$ , how do you find the positive interval of unity length, in which these two functions intersect?

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Given the functions f(x)=x2−4f(x)=x^2 -4 and g(x)=2x−3g(x)= 2x -3, how do you find the positive interval of unity length, in which these two functions intersect?

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Given the functions $f (x) = x^{2} - 4$ $f(x)=x^2 -4$ and $g (x) = 2 x - 3$ $g(x)= 2x -3$ , how do you find the positive interval of unity length, in which these two functions intersect?