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Question Number 192132 by Red1ight last updated on 09/May/23
What is the nearest point in f(x) to (5,2)  where f(x)=−0.5x^2 +3
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{point}\:\mathrm{in}\:{f}\left({x}\right)\:\mathrm{to}\:\left(\mathrm{5},\mathrm{2}\right) \\ $$$$\mathrm{where}\:{f}\left({x}\right)=−\mathrm{0}.\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3} \\ $$
Answered by mehdee42 last updated on 09/May/23
according to the diagram:  let :H(h,f(h))    f′(x)=−x ⇒m_l_1  =−h  &  m_l_2  =((2−f(h))/(5−h))  m_l_1  ×m_l_2  =−1⇒((2−0.5h^2 −3)/(5−h))×(−h)=−1⇒h=((10))^(1/3)   ⇒H(((10))^(1/3)  , 3−0.5((100))^(1/3) ) ✓
$${according}\:{to}\:{the}\:{diagram}: \\ $$$${let}\::{H}\left({h},{f}\left({h}\right)\right)\:\: \\ $$$${f}'\left({x}\right)=−{x}\:\Rightarrow{m}_{{l}_{\mathrm{1}} } =−{h}\:\:\&\:\:{m}_{{l}_{\mathrm{2}} } =\frac{\mathrm{2}−{f}\left({h}\right)}{\mathrm{5}−{h}} \\ $$$${m}_{{l}_{\mathrm{1}} } ×{m}_{{l}_{\mathrm{2}} } =−\mathrm{1}\Rightarrow\frac{\mathrm{2}−\mathrm{0}.\mathrm{5}{h}^{\mathrm{2}} −\mathrm{3}}{\mathrm{5}−{h}}×\left(−{h}\right)=−\mathrm{1}\Rightarrow{h}=\sqrt[{\mathrm{3}}]{\mathrm{10}} \\ $$$$\Rightarrow{H}\left(\sqrt[{\mathrm{3}}]{\mathrm{10}}\:,\:\mathrm{3}−\mathrm{0}.\mathrm{5}\sqrt[{\mathrm{3}}]{\mathrm{100}}\right)\:\checkmark \\ $$
Commented by mehdee42 last updated on 09/May/23
Commented by Red1ight last updated on 09/May/23
  why m_l_1  =−h?
$$ \\ $$$$\mathrm{why}\:{m}_{{l}_{\mathrm{1}} } =−{h}? \\ $$
Commented by mehdee42 last updated on 10/May/23
the line  l_1  is tangent to the corve at point H.  so its slope is equal  to the value of the   derivative of the function per length of   the tangent point . that′s mean m_l_1  =f ′(h)=−h  ok?
$${the}\:{line}\:\:{l}_{\mathrm{1}} \:{is}\:{tangent}\:{to}\:{the}\:{corve}\:{at}\:{point}\:{H}. \\ $$$${so}\:{its}\:{slope}\:{is}\:{equal}\:\:{to}\:{the}\:{value}\:{of}\:{the}\: \\ $$$${derivative}\:{of}\:{the}\:{function}\:{per}\:{length}\:{of}\: \\ $$$${the}\:{tangent}\:{point}\:.\:{that}'{s}\:{mean}\:{m}_{{l}_{\mathrm{1}} } ={f}\:'\left({h}\right)=−{h} \\ $$$${ok}? \\ $$
Commented by Red1ight last updated on 09/May/23
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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