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Question Number 69355 by cesar.marval.larez@gmail.com last updated on 22/Sep/19
  Find the maximun and minimum   values of the function f(x)=x^2 −8x+7  and sketch its graph
$$ \\ $$$${Find}\:{the}\:{maximun}\:{and}\:{minimum}\: \\ $$$${values}\:{of}\:{the}\:{function}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{7} \\ $$$${and}\:{sketch}\:{its}\:{graph}\: \\ $$
Answered by MJS last updated on 22/Sep/19
min/max at f′(x)=0  to sketch it we also need the zeros    zeros  x^2 −8x+7=0  x=−(p/2)±(√((p^2 /4)−q))  p=−8; q=7  x=4±(√(16−7))  x_1 =1  x_2 =7    min  f′(x)=2x−8=0 ⇒ x=4  f′′(x)=2>0 ⇒ minimum at  ((4),((−9)) )    now sketch it
$$\mathrm{min}/\mathrm{max}\:\mathrm{at}\:{f}'\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{to}\:\mathrm{sketch}\:\mathrm{it}\:\mathrm{we}\:\mathrm{also}\:\mathrm{need}\:\mathrm{the}\:\mathrm{zeros} \\ $$$$ \\ $$$$\mathrm{zeros} \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{7}=\mathrm{0} \\ $$$${x}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$$${p}=−\mathrm{8};\:{q}=\mathrm{7} \\ $$$${x}=\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{7}} \\ $$$${x}_{\mathrm{1}} =\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\mathrm{7} \\ $$$$ \\ $$$$\mathrm{min} \\ $$$${f}'\left({x}\right)=\mathrm{2}{x}−\mathrm{8}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{4} \\ $$$${f}''\left({x}\right)=\mathrm{2}>\mathrm{0}\:\Rightarrow\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{\mathrm{4}}\\{−\mathrm{9}}\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{sketch}\:\mathrm{it} \\ $$

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