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Question Number 116776 by abony1303 last updated on 06/Oct/20

Given function:  f(x)= { ((2x^2 +1,  ∣x∣<3)),((5x−1,  ∣x∣≥3)) :}  Find out: f(x^2 +7)=?    A)5x^2 −34     B)2x^2 +8   C)5x^2 +36  D)5x^2 +34    E)2(x^2 +7)^2 +1  Please help

$$\mathrm{Given}\:\mathrm{function}: \\ $$ $${f}\left({x}\right)=\begin{cases}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1},\:\:\mid{x}\mid<\mathrm{3}}\\{\mathrm{5x}−\mathrm{1},\:\:\mid{x}\mid\geqslant\mathrm{3}}\end{cases} \\ $$ $$\mathrm{Find}\:\mathrm{out}:\:{f}\left({x}^{\mathrm{2}} +\mathrm{7}\right)=? \\ $$ $$ \\ $$ $$\left.\mathrm{A}\left.\right)\left.\mathrm{5x}^{\mathrm{2}} −\mathrm{34}\:\:\:\:\:\mathrm{B}\right)\mathrm{2x}^{\mathrm{2}} +\mathrm{8}\:\:\:\mathrm{C}\right)\mathrm{5x}^{\mathrm{2}} +\mathrm{36} \\ $$ $$\left.\mathrm{D}\left.\right)\mathrm{5x}^{\mathrm{2}} +\mathrm{34}\:\:\:\:\mathrm{E}\right)\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7}\right)^{\mathrm{2}} +\mathrm{1} \\ $$ $$\mathrm{Please}\:\mathrm{help} \\ $$

Answered by floor(10²Eta[1]) last updated on 06/Oct/20

∣x^2 +7∣≥7>3⇒f(x^2 +7)=5(x^2 +7)−1  =5x^2 +34

$$\mid\mathrm{x}^{\mathrm{2}} +\mathrm{7}\mid\geqslant\mathrm{7}>\mathrm{3}\Rightarrow\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7}\right)=\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7}\right)−\mathrm{1} \\ $$ $$=\mathrm{5x}^{\mathrm{2}} +\mathrm{34} \\ $$

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