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Question Number 60439 by ajfour last updated on 21/May/19

Commented by ajfour last updated on 21/May/19

$$\mathrm{Find}\mathrm{length}\mathrm{AB}.\mathrm{The}\mathrm{curve}\mathrm{is}\mathrm{cubic}.$$

Answered by tanmay last updated on 21/May/19

Commented by ajfour last updated on 21/May/19

$$\mathrm{thanks}\mathrm{Tanmay}\mathrm{Sir}.$$

Commented by tanmay last updated on 21/May/19

$$mostwelcomesir...$$

Answered by mr W last updated on 21/May/19

$$f\left(x\right)=a(x-4)(x-2)(x+1)$$$$f\left(0\right)=a(-4)(-2)\left(1\right)=1$$$$\Rightarrow a=\frac{1}{8}$$$$\Rightarrow f\left(x\right)=\frac{1}{8}(x-4)(x-2)(x+1)$$$$\Rightarrow f\left(x\right)=\frac{1}{8}(x^3-5x^2+2x+8)$$$$f^{\prime }\left(x\right)=\frac{1}{8}(3x^2-10x+2)$$$$A(h,k)$$$$k=\frac{1}{8}(h^3-5h^2+2h+8)$$$$m_A=\frac{1}{8}(3h^2-10h+2)$$$$eqn.ofAB:$$$$y-k=m_A(x-h)$$$$B(4,0)$$$$0-k=m_A(4-h)$$$$0-\frac{1}{8}(h^3-5h^2+2h+8)=\frac{1}{8}(3h^2-10h+2)(4-h)$$$$-h^3+5h^2-2h-8=-3h^3+22h^2-42h+8$$$$\Rightarrow 2h^3-17h^2+40h-16=0$$$$\Rightarrow h=\frac{1}{2}$$$$\Rightarrow k=\frac{1}{8}(\frac{1}{8}-\frac{5}{4}+1+8)=\frac{63}{64}$$$$\Rightarrow AB=\sqrt{{(4-\frac{1}{2})}^2+{\left(\frac{63}{64}\right)}^2}=\frac{7\sqrt{1105}}{64}=3.64$$

Commented by ajfour last updated on 21/May/19

$$\mathrm{thanks}\mathrm{sir}.$$

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