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Question Number 200168 by hardmath last updated on 15/Nov/23

If  f(x) = 2^x  + 86  and  g(x) = 3x^2  + x − 4  Then find:  g[f^(−1) (g(14))] = ?

$$\mathrm{If}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{86}\:\:\mathrm{and}\:\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{x}\:−\:\mathrm{4} \\ $$$$\mathrm{Then}\:\mathrm{find}:\:\:\mathrm{g}\left[\mathrm{f}^{−\mathrm{1}} \left(\mathrm{g}\left(\mathrm{14}\right)\right)\right]\:=\:? \\ $$

Commented by jazeee last updated on 15/Nov/23

  Solution:  We need to find f^(−1) (x) first.  y = 2^x  + 86  x = 2^x  + 86  x − 86 = 2^y   ln (x − 86) = y ∗ ln (2)  ((ln (x − 86))/(ln (2))) = y  f^(−1)  (x) = ((ln (x −86))/(ln(2)))    Now, substitute 14 in g(x).  g(14) = 3(14)^2  + (14) − 4  g(14) = 3(196) + 10  g(14) = 588 + 10  g(14) = 598    Next is to substitute 598 in f^(−1)  (x).  f^(−1)  (598) = ((ln (598 −86))/(ln (2)))  f^(−1)  (598) = ((ln (512))/(ln (2)))  note: 512 = 2^9   f^(−1)  (598) = ((9 ln(2))/(ln (2)))  f^(−1)  (598) = 9    Lastly, substitute 9 in g(x).  g(9) = 3(9)^2  + (9) − 4  g(9) = 3(81) + 5  g(9) = 243 + 5  g(9) = 248    Answer: 248.

$$ \\ $$$${Solution}: \\ $$$${We}\:{need}\:{to}\:{find}\:{f}^{−\mathrm{1}} \left({x}\right)\:{first}. \\ $$$${y}\:=\:\mathrm{2}^{{x}} \:+\:\mathrm{86} \\ $$$${x}\:=\:\mathrm{2}^{{x}} \:+\:\mathrm{86} \\ $$$${x}\:−\:\mathrm{86}\:=\:\mathrm{2}^{{y}} \\ $$$${ln}\:\left({x}\:−\:\mathrm{86}\right)\:=\:{y}\:\ast\:{ln}\:\left(\mathrm{2}\right) \\ $$$$\frac{{ln}\:\left({x}\:−\:\mathrm{86}\right)}{{ln}\:\left(\mathrm{2}\right)}\:=\:{y} \\ $$$${f}^{−\mathrm{1}} \:\left({x}\right)\:=\:\frac{{ln}\:\left({x}\:−\mathrm{86}\right)}{{ln}\left(\mathrm{2}\right)} \\ $$$$ \\ $$$${Now},\:{substitute}\:\mathrm{14}\:{in}\:{g}\left({x}\right). \\ $$$${g}\left(\mathrm{14}\right)\:=\:\mathrm{3}\left(\mathrm{14}\right)^{\mathrm{2}} \:+\:\left(\mathrm{14}\right)\:−\:\mathrm{4} \\ $$$${g}\left(\mathrm{14}\right)\:=\:\mathrm{3}\left(\mathrm{196}\right)\:+\:\mathrm{10} \\ $$$${g}\left(\mathrm{14}\right)\:=\:\mathrm{588}\:+\:\mathrm{10} \\ $$$${g}\left(\mathrm{14}\right)\:=\:\mathrm{598} \\ $$$$ \\ $$$${Next}\:{is}\:{to}\:{substitute}\:\mathrm{598}\:{in}\:{f}^{−\mathrm{1}} \:\left({x}\right). \\ $$$${f}^{−\mathrm{1}} \:\left(\mathrm{598}\right)\:=\:\frac{{ln}\:\left(\mathrm{598}\:−\mathrm{86}\right)}{{ln}\:\left(\mathrm{2}\right)} \\ $$$${f}^{−\mathrm{1}} \:\left(\mathrm{598}\right)\:=\:\frac{{ln}\:\left(\mathrm{512}\right)}{{ln}\:\left(\mathrm{2}\right)} \\ $$$${note}:\:\mathrm{512}\:=\:\mathrm{2}^{\mathrm{9}} \\ $$$${f}^{−\mathrm{1}} \:\left(\mathrm{598}\right)\:=\:\frac{\mathrm{9}\:{ln}\left(\mathrm{2}\right)}{{ln}\:\left(\mathrm{2}\right)} \\ $$$${f}^{−\mathrm{1}} \:\left(\mathrm{598}\right)\:=\:\mathrm{9} \\ $$$$ \\ $$$${Lastly},\:{substitute}\:\mathrm{9}\:{in}\:{g}\left({x}\right). \\ $$$${g}\left(\mathrm{9}\right)\:=\:\mathrm{3}\left(\mathrm{9}\right)^{\mathrm{2}} \:+\:\left(\mathrm{9}\right)\:−\:\mathrm{4} \\ $$$${g}\left(\mathrm{9}\right)\:=\:\mathrm{3}\left(\mathrm{81}\right)\:+\:\mathrm{5} \\ $$$${g}\left(\mathrm{9}\right)\:=\:\mathrm{243}\:+\:\mathrm{5} \\ $$$${g}\left(\mathrm{9}\right)\:=\:\mathrm{248} \\ $$$$ \\ $$$${Answer}:\:\mathrm{248}. \\ $$

Commented by jazeee last updated on 15/Nov/23

  ∗ x = 2^y  + 86 ∗ (second step)

$$ \\ $$$$\ast\:{x}\:=\:\mathrm{2}^{{y}} \:+\:\mathrm{86}\:\ast\:\left({second}\:{step}\right) \\ $$

Answered by Sutrisno last updated on 15/Nov/23

g(14)=3(14)^2 +14−4=598  f(x)=2^x +86→x=f^(−1) (2^x +86)  2^x +86=598  2^x =512→x=9  g(9)=3(9)^2 +9−4=248  ∴g(f^(−1) (g(14)))=248

$${g}\left(\mathrm{14}\right)=\mathrm{3}\left(\mathrm{14}\right)^{\mathrm{2}} +\mathrm{14}−\mathrm{4}=\mathrm{598} \\ $$$${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{86}\rightarrow{x}={f}^{−\mathrm{1}} \left(\mathrm{2}^{{x}} +\mathrm{86}\right) \\ $$$$\mathrm{2}^{{x}} +\mathrm{86}=\mathrm{598} \\ $$$$\mathrm{2}^{{x}} =\mathrm{512}\rightarrow{x}=\mathrm{9} \\ $$$${g}\left(\mathrm{9}\right)=\mathrm{3}\left(\mathrm{9}\right)^{\mathrm{2}} +\mathrm{9}−\mathrm{4}=\mathrm{248} \\ $$$$\therefore{g}\left({f}^{−\mathrm{1}} \left({g}\left(\mathrm{14}\right)\right)\right)=\mathrm{248} \\ $$

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