A curve has equation $$\(y=x^{3} \mathrm{e}^{0.2 x}\)$$ where $$\(x \geqslant 0\)$$. At the point $$\(P\)$$ on the curve, the gradient of the curve is 15 . Use the equation in part (a) to show by calculation that the $$\(x\)$$-coordinate of $$\(P\)$$ lies between $$\(1.7\) and \(1.8\)$$ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No:9709_m20_qp_22 Year:2020 Question No:6(b)
Answer:
Consider sign of \(x-\sqrt{\frac{75 \mathrm{e}^{-0.2 x}}{15+x}}\) or equivalent for \(1.7\) and \(1.8\)
Obtain \(-0.08 \ldots\) and \(0.03 \ldots\) or equivalents and justify conclusion
Obtain \(-0.08 \ldots\) and \(0.03 \ldots\) or equivalents and justify conclusion
Knowledge points:
2.6.1 locate approximately a root of an equation, by means of graphical considerations and/or searching for a sign change
Solution:
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