Question

F(t)=4t-5,g(t)=t^2,h(t)=1/t find thevalue for fohog(4)

Answer 1

Answer:

$-\frac{19}{4}$

Step-by-step explanation:

Given functions,

$f(t)=4t-5------(1)$,

$g(t)=t^2------(2)$,

$h(t)=\frac{1}{t}-----(3)$

∵ hog(t) = h(g(t)) = $h(t^2)$  ( from equation (2) ),

$\implies hog(t) = \frac{1}{t^2}----(4)$     ( from equation (3) ),

Thus,

fohog(t) = f(hog(t))

$\implies fohog(t) = f(\frac{1}{t^2})$ ( from equation (4) ),

$=4(\frac{1}{t^2})-5$  ( from equation (1) )

Hence,

$fohog(4) = 4(\frac{1}{16})-5=\frac{1}{4}-5=\frac{1-20}{4}=-\frac{19}{4}$

Answer 2

Answer:

$\red { Value \: of \: fohog(4) }\green { =\frac{-19}{4}}$

Step-by-step explanation:

f(t) = 4t - 5 ,

g(t) = t² ,

$h(t) = \frac{1}{t}$

$fohog(x)$

$= f[h(g(x))]$

$=f[h(t^{2})]$

$=f[\frac{1}{t^{2}}]\\=4\times \frac{1}{t^{2}} - 5 \: ---(1)$

$Now , \red { Value \: of \: fohog(4) }$

$= 4\times \frac{1}{4^{2}} - 5$

$= \frac{4}{16} - 5\\= \frac{1}{4} - 5\\=\frac{1-20}{4}\\=\green { \frac{-19}{4}}$

Therefore.,

$\red { Value \: of \: fohog(4) }\green { =\frac{-19}{4}}$

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