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Question Number 88918 by mathocean1 last updated on 13/Apr/20

Commented by mathocean1 last updated on 13/Apr/20

Commented by mathocean1 last updated on 13/Apr/20

$$\mathrm{I}\mathrm{precise}\mathrm{that}\mathrm{f}\mathrm{\exists }\mathrm{\forall }\mathrm{x}\ne \frac{1}{2}$$

Commented by mathmax by abdo last updated on 13/Apr/20

$$howtofindfwiththisvariation...!$$

Commented by mathocean1 last updated on 14/Apr/20

$$\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{also}\mathrm{know}$$

Commented by MJS last updated on 14/Apr/20

$$f^{\prime }\left(x\right)=0\mathrm{at}x=-1\wedge x=3$$$$\Rightarrow (x+1)(x-3)$$$$f^{\prime }\left(x\right)\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{at}x=\frac{1}{2}\mathrm{and}\mathrm{the}\mathrm{sign}$$$${\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{change}$$$$\Rightarrow \frac{1}{{(x-\frac{1}{2})}^2}$$$$\Rightarrow$$$$f^{\prime }\left(x\right)=\frac{(x+1)(x-3)}{{(x-\frac{1}{2})}^2}\times c_1$$$$f\left(x\right)=\int f^{\prime }\left(x\right)dx=c_1(x+\frac{15}{4(x-\frac{1}{2})}-\ln \mid 2x-1\mid )+c_2$$$$\mathrm{now}\mathrm{calculate}{c}_1,c_2\mathrm{by}\mathrm{solving}$$$$\{\begin{array}{l}f(-1)=\frac{1}{2}\\ f\left(3\right)=2\end{array}$$$$\Rightarrow$$$$c_1=\frac{3}{2(8-\ln 5-\ln 3)}\wedge c_2=\frac{37-2\ln 5-8\ln 3}{4(8-\ln 5-\ln 3)}$$

Commented by mathocean1 last updated on 14/Apr/20

$$\mathrm{Please}\mathrm{i}\mathrm{dont}\mathrm{know}\mathrm{these}\mathrm{characters}:$$$$\ln ;\int \mathrm{and}\mathrm{d}x.$$$$i^{\prime }\mathrm{m}\mathrm{till}\mathrm{learner}:)!$$

Commented by MJS last updated on 14/Apr/20

$$\mathrm{how}\mathrm{come}\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{solve}\mathrm{a}\mathrm{question}\mathrm{like}$$$$\mathrm{this}\mathrm{without}\mathrm{knowing}\mathrm{these}\mathrm{symbols}???$$$$\mathrm{I}\mathrm{cannot}\mathrm{explain}\mathrm{this}\mathrm{within}\mathrm{a}\mathrm{comment},\mathrm{it}$$$$\mathrm{needs}\mathrm{teaching}\mathrm{for}\mathrm{months}$$

Commented by mathocean1 last updated on 15/Apr/20

$$\mathrm{alright}.$$

Commented by MJS last updated on 15/Apr/20

$$\mathrm{sorry}\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{want}\mathrm{to}\mathrm{be}\mathrm{impolite}$$$$\mathrm{I}\mathrm{cannot}\mathrm{see}\mathrm{an}\mathrm{easier}\mathrm{path}\mathrm{to}\mathrm{solve}\mathrm{this}$$$$\mathrm{problem}.\mathrm{in}\mathrm{Austria},\mathrm{we}\mathrm{have}12\mathrm{years}\mathrm{of}$$$$\mathrm{school},\mathrm{you}\mathrm{learn}‘‘{\ln }^{″}\mathrm{in}\mathrm{the}{10}^{\mathrm{th}}\mathrm{year}\mathrm{and}$$$$‘‘\int {dx}^{″}\mathrm{in}\mathrm{the}{12}^{\mathrm{th}}\mathrm{year}...$$$$\mathrm{just}\mathrm{to}\mathrm{give}\mathrm{you}\mathrm{a}\mathrm{picture}$$

Commented by mathocean1 last updated on 08/Jun/20

$$thankssir$$

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