<LINK href="file:///C:%5CUsers%5CMelissa%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_filelist.xml" rel=File-List><LINK href="file:///C:%5CUsers%5CMelissa%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_themedata.thmx" rel=themeData><LINK href="file:///C:%5CUsers%5CMelissa%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_colorschememapping.xml" rel=colorSchemeMapping><STYLE> <!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format
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<O>
</O>ii) Let y = e<SUP>x</SUP> - e<SUP>-x</SUP>
Swap x and y values:
x =e<SUP>y</SUP> - e<SUP>-y </SUP>=<SUP> </SUP>e<SUP>y</SUP> - 1/e<SUP>y<O></O></SUP>
e<SUP>y</SUP>x = e<SUP>2y</SUP> - 1
e<SUP>2y</SUP> - e<SUP>y</SUP>x – 1 = 0
Now this is a quadratic with e<SUP>y</SUP>
Therefore e<SUP>y </SUP>= (x +- sqrt(x<SUP>2</SUP> +4))/2
But e<SUP>y</SUP> is always greater than 0
So e<SUP>y </SUP>= (x + sqrt(x<SUP>2</SUP> +4))/2
Therefore y = log<SUB>e</SUB>(x + sqrt(x<SUP>2</SUP> +4))/2
f<SUP>-I</SUP>(x) = log<SUB>e</SUB>(x + sqrt(x<SUP>2</SUP> +4))/2
<O></O>
iii) Sub 5 into the equation in part ii)
f<SUP>-I</SUP>(x) = log<SUB>e</SUB>(5 + sqrt(25 +4))/2
= 1.65 (2dp)
Therefore x = 1.65 (2dp)