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Question-179338

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Question Number 179338 by mathlove last updated on 28/Oct/22
Answered by Acem last updated on 28/Oct/22
= 20× − (5/3) = − ((100)/3) if 20 was num. of relation, the answer is − (5/3)
$$=\:\mathrm{20}×\:−\:\frac{\mathrm{5}}{\mathrm{3}}\:=\:−\:\frac{\mathrm{100}}{\mathrm{3}}\: \\ $$$${if}\:\mathrm{20}\:{was}\:{num}.\:{of}\:{relation}, \\ $$$$\:{the}\:{answer}\:{is}\:−\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Commented by mathlove last updated on 28/Oct/22
way f^(−1) (5)=1 ??
$${way}\:{f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{1}\:?? \\ $$
Commented by Acem last updated on 30/Oct/22
way f^(−1) (5)=1 ?? Hello friend, you solved f^( −1) (0) correctly as −4 now −4 for which x or 1−x, sure for 1−x 1−x= −4⇒ x=5 i.e. f^( −1) (0)= 5 The same thing for f^( −1) (5) we get 0 and it′s for 1−x= 0 ⇒ x= 1 i.e. f^( −1) (5)= 1
$${way}\:{f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{1}\:?? \\ $$$${Hello}\:{friend},\:{you}\:{solved}\:{f}^{\:−\mathrm{1}} \left(\mathrm{0}\right)\:{correctly}\:{as}\:−\mathrm{4} \\ $$$$\:{now}\:−\mathrm{4}\:{for}\:{which}\:{x}\:{or}\:\mathrm{1}−{x},\:{sure}\:{for}\:\mathrm{1}−{x} \\ $$$$\:\mathrm{1}−{x}=\:−\mathrm{4}\Rightarrow\:{x}=\mathrm{5}\:{i}.{e}.\:{f}^{\:−\mathrm{1}} \left(\mathrm{0}\right)=\:\mathrm{5} \\ $$$$ \\ $$$${The}\:{same}\:{thing}\:{for}\:{f}^{\:−\mathrm{1}} \left(\mathrm{5}\right)\:{we}\:{get}\:\mathrm{0}\:{and}\:{it}'{s}\:{for} \\ $$$$\:\mathrm{1}−{x}=\:\mathrm{0}\:\Rightarrow\:{x}=\:\mathrm{1}\:{i}.{e}.\:\boldsymbol{{f}}^{\:−\mathrm{1}} \left(\mathrm{5}\right)=\:\mathrm{1} \\ $$$$\: \\ $$
Commented by Acem last updated on 30/Oct/22
Now try to find the value of f(1−x) for x= 1 ⇒ f(0)= 5
$${Now}\:{try}\:{to}\:{find}\:{the}\:{value}\:{of}\:{f}\left(\mathrm{1}−{x}\right)\:{for}\:{x}=\:\mathrm{1} \\ $$$$\:\Rightarrow\:{f}\left(\mathrm{0}\right)=\:\mathrm{5} \\ $$
Commented by Acem last updated on 30/Oct/22
When we write f(0) or f(7) that doesn′t mean that x=0 neither x= 7 but means 1−x=0, 1−x= 7
$${When}\:{we}\:{write}\:{f}\left(\mathrm{0}\right)\:{or}\:{f}\left(\mathrm{7}\right)\:{that}\:{doesn}'{t}\:{mean}\:{that} \\ $$$$\:{x}=\mathrm{0}\:{neither}\:{x}=\:\mathrm{7}\:{but}\:{means}\:\mathrm{1}−{x}=\mathrm{0},\:\mathrm{1}−{x}=\:\mathrm{7} \\ $$
Answered by mathlove last updated on 28/Oct/22
chek this f(7)=f(1−7)=f(−6)=−2 f(−4)=f(1−(−4))=f(5)=8 f^(−1) (5)=0 and f^(−1) (0)=5 ((f(7)−f(−4))/(f^(−1) (5)+f^(−1) (0)))=((−2−8)/5)=((−10)/5)=−2
$${chek}\:{this} \\ $$$${f}\left(\mathrm{7}\right)={f}\left(\mathrm{1}−\mathrm{7}\right)={f}\left(−\mathrm{6}\right)=−\mathrm{2} \\ $$$${f}\left(−\mathrm{4}\right)={f}\left(\mathrm{1}−\left(−\mathrm{4}\right)\right)={f}\left(\mathrm{5}\right)=\mathrm{8} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{0}\:\:\:\:{and}\:\:\:\:{f}^{−\mathrm{1}} \left(\mathrm{0}\right)=\mathrm{5} \\ $$$$\frac{{f}\left(\mathrm{7}\right)−{f}\left(−\mathrm{4}\right)}{{f}^{−\mathrm{1}} \left(\mathrm{5}\right)+{f}^{−\mathrm{1}} \left(\mathrm{0}\right)}=\frac{−\mathrm{2}−\mathrm{8}}{\mathrm{5}}=\frac{−\mathrm{10}}{\mathrm{5}}=−\mathrm{2} \\ $$
Commented by Acem last updated on 28/Oct/22
f^( −1) (5)= 0 ⇒ x= 1 not 0 ; F(x)= f(1−x) so the answer is ((−10)/(1+5))= − (5/3)
$${f}^{\:−\mathrm{1}} \left(\mathrm{5}\right)=\:\mathrm{0}\:\Rightarrow\:{x}=\:\mathrm{1}\:{not}\:\mathrm{0}\:;\:{F}\left({x}\right)=\:{f}\left(\mathrm{1}−{x}\right) \\ $$$$\:{so}\:{the}\:{answer}\:{is}\:\frac{−\mathrm{10}}{\mathrm{1}+\mathrm{5}}=\:−\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 28/Oct/22
f(7)=f(1−x)∣_(x=−6) =−2 f(−4)=f(1−x)∣_(x=5) =8 f^(−1) (5)=1 f^(−1) (0)=5 ((f(7)−f(−4))/(f^(−1) (5)+f^(−1) (0)))=((−2−8)/(1+5))=−(5/3) ✓
$${f}\left(\mathrm{7}\right)={f}\left(\mathrm{1}−{x}\right)\mid_{{x}=−\mathrm{6}} =−\mathrm{2} \\ $$$${f}\left(−\mathrm{4}\right)={f}\left(\mathrm{1}−{x}\right)\mid_{{x}=\mathrm{5}} =\mathrm{8} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{1}\:\: \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{0}\right)=\mathrm{5}\:\: \\ $$$$\frac{{f}\left(\mathrm{7}\right)−{f}\left(−\mathrm{4}\right)}{{f}^{−\mathrm{1}} \left(\mathrm{5}\right)+{f}^{−\mathrm{1}} \left(\mathrm{0}\right)}=\frac{−\mathrm{2}−\mathrm{8}}{\mathrm{1}+\mathrm{5}}=−\frac{\mathrm{5}}{\mathrm{3}}\:\checkmark \\ $$
Commented by mr W last updated on 29/Oct/22
Commented by mr W last updated on 29/Oct/22
then we can get directly: f(7)=−2, f(−4)=8 f^(−1) (5)=1, f^(−1) (6)=−2
$${then}\:{we}\:{can}\:{get}\:{directly}: \\ $$$${f}\left(\mathrm{7}\right)=−\mathrm{2},\:{f}\left(−\mathrm{4}\right)=\mathrm{8} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{1},\:{f}^{−\mathrm{1}} \left(\mathrm{6}\right)=−\mathrm{2} \\ $$
Commented by mr W last updated on 29/Oct/22
we can also transform the graph of f(1−x) into the graph of f(x) by adjusting the coordinate axes.
$${we}\:{can}\:{also}\:{transform}\:{the}\:{graph} \\ $$$${of}\:{f}\left(\mathrm{1}−{x}\right)\:{into}\:{the}\:{graph}\:{of}\:{f}\left({x}\right) \\ $$$${by}\:{adjusting}\:{the}\:{coordinate}\:{axes}. \\ $$
Commented by Acem last updated on 28/Oct/22
It′s multiplied by 20 i.e. −((100)/3) , anyway if 20 was num. of relation that′s not problem
$${It}'{s}\:{multiplied}\:{by}\:\mathrm{20}\:{i}.{e}.\:−\frac{\mathrm{100}}{\mathrm{3}}\:,\:{anyway}\:{if}\:\mathrm{20} \\ $$$$\:{was}\:{num}.\:{of}\:{relation}\:{that}'{s}\:{not}\:{problem} \\ $$
Commented by mr W last updated on 28/Oct/22
i think 20 is just the question numbering.
$${i}\:{think}\:\mathrm{20}\:{is}\:{just}\:{the}\:{question}\: \\ $$$${numbering}. \\ $$
Commented by Acem last updated on 28/Oct/22
Yes haha it′s a common mistake in writing though we have a lot of other things “ 20) , 20• ”
$${Yes}\:{haha}\:{it}'{s}\:{a}\:{common}\:{mistake}\:{in}\:{writing} \\ $$$$\left.\:{though}\:{we}\:{have}\:\:{a}\:{lot}\:{of}\:{other}\:{things}\:“\:\mathrm{20}\right)\:,\:\mathrm{20}\bullet\:'' \\ $$
Commented by mathlove last updated on 28/Oct/22
yes 20 Q number
$${yes}\:\mathrm{20}\:{Q}\:{number} \\ $$
Commented by mathlove last updated on 28/Oct/22
thanks to all
$${thanks}\:{to}\:{all} \\ $$
Commented by mathlove last updated on 28/Oct/22
i do′nt understand f^(−1) (5)=^? 1
$${i}\:{do}'{nt}\:{understand}\:{f}^{−\mathrm{1}} \left(\mathrm{5}\right)\overset{?} {=}\mathrm{1} \\ $$
Commented by mr W last updated on 28/Oct/22
f^(−1) (5) means: what is x if f(x)=5? we see from the graph: at x=0: f(1−x)=5, i.e. f(1)=5 therefore f^(−1) (5)=1.
$${f}^{−\mathrm{1}} \left(\mathrm{5}\right)\:{means}:\:{what}\:{is}\:{x}\:{if}\:{f}\left({x}\right)=\mathrm{5}? \\ $$$$ \\ $$$${we}\:{see}\:{from}\:{the}\:{graph}: \\ $$$${at}\:{x}=\mathrm{0}:\:{f}\left(\mathrm{1}−{x}\right)=\mathrm{5},\:{i}.{e}.\:{f}\left(\mathrm{1}\right)=\mathrm{5} \\ $$$${therefore}\:{f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{1}. \\ $$
Commented by mr W last updated on 28/Oct/22
similarly we see: at x=−4: f(1−x)=0, i.e. f(5)=0, therefore f^(−1) (0)=5.
$${similarly}\:{we}\:{see}: \\ $$$${at}\:{x}=−\mathrm{4}:\:{f}\left(\mathrm{1}−{x}\right)=\mathrm{0},\:{i}.{e}.\:{f}\left(\mathrm{5}\right)=\mathrm{0}, \\ $$$${therefore}\:{f}^{−\mathrm{1}} \left(\mathrm{0}\right)=\mathrm{5}. \\ $$
Commented by mathlove last updated on 28/Oct/22
thanks mr W
$${thanks}\:{mr}\:{W} \\ $$
Commented by mathlove last updated on 30/Oct/22
thanks so much mr
$${thanks}\:{so}\:{much}\:{mr} \\ $$

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