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# Find the largest value of x where the plots of ​ intersect.

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Find the largest value of x where the plots of $$f(x) = - \frac{2x+5}{x+3} \text{ and } g(x) = 4\cdot \frac{x+1}{x-4}$$
intersect.

waffles  Mar 9, 2018
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Find the largest value of x where the plots of

$$\displaystyle f(x) = - \frac{2x+5}{x+3} \qquad \text{ and } \qquad g(x) = 4\cdot \frac{x+1}{x-4}$$

f(x) = - \frac{2x+5}{x+3} \text{ and } g(x) = 4\cdot \frac{x+1}{x-4}
intersect.

$$\begin{array}{|rcll|} \hline - \dfrac{2x+5}{x+3} &=& 4\cdot \frac{x+1}{x-4} \\\\ -(2x+5)(x-4) &=& 4\cdot (x+1)(x+3) \\ 4\cdot (x+1)(x+3) &=& -(2x+5)(x-4) \\ 4\cdot (x^2+3x+x+3) &=& -(2x^2 -8x +5x -20) \\ 4\cdot (x^2+4x+3) &=& -(2x^2 -3x -20) \\ 4x^2+16x+12 &=& -2x^2 +3x +20 \\ 4x^2+2x^2+16x -3x+12-20 &=& 0 \\ 6x^2+13x-8 &=& 0 \\\\ x_{1,2} &=& \dfrac{ -13\pm \sqrt{13^2-4\cdot 6 \cdot (-8) } } {2\cdot 6} \\\\ &=& \dfrac{ -13\pm \sqrt{169+192 } } {12} \\\\ &=& \dfrac{ -13\pm \sqrt{361} } {12} \\\\ &=& \dfrac{ -13\pm 19 } {12} \\\\ x_{\text{max}} &=& \dfrac{ -13+ 19 } {12} \\\\ &=& \dfrac{ 6 } {12} \\\\ &=& \dfrac{ 1 } {2} \\\\ \mathbf{x_{\text{max}}} &\mathbf{=}&\mathbf{ 0.5 } \\ \hline \end{array}$$

heureka  Mar 9, 2018

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