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

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

Commented by jazeee last updated on 15/Nov/23

$$$$$$Solution:$$$$Weneedtofind{f}^{-1}\left(x\right)first.$$$$y=2^x+86$$$$x=2^x+86$$$$x-86=2^y$$$$ln(x-86)=y\ast ln\left(2\right)$$$$\frac{ln(x-86)}{ln\left(2\right)}=y$$$$f^{-1}\left(x\right)=\frac{ln(x-86)}{ln\left(2\right)}$$$$$$$$Now,substitute14ing\left(x\right).$$$$g\left(14\right)=3{\left(14\right)}^2+\left(14\right)-4$$$$g\left(14\right)=3\left(196\right)+10$$$$g\left(14\right)=588+10$$$$g\left(14\right)=598$$$$$$$$Nextistosubstitute598in{f}^{-1}\left(x\right).$$$$f^{-1}\left(598\right)=\frac{ln(598-86)}{ln\left(2\right)}$$$$f^{-1}\left(598\right)=\frac{ln\left(512\right)}{ln\left(2\right)}$$$$note:512=2^9$$$$f^{-1}\left(598\right)=\frac{9ln\left(2\right)}{ln\left(2\right)}$$$$f^{-1}\left(598\right)=9$$$$$$$$Lastly,substitute9ing\left(x\right).$$$$g\left(9\right)=3{\left(9\right)}^2+\left(9\right)-4$$$$g\left(9\right)=3\left(81\right)+5$$$$g\left(9\right)=243+5$$$$g\left(9\right)=248$$$$$$$$Answer:248.$$

Commented by jazeee last updated on 15/Nov/23

$$$$$$\ast x=2^y+86\ast \left(secondstep\right)$$

Answered by Sutrisno last updated on 15/Nov/23

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

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