Let $f, g, h$ are functions defined by $f (x) = x; g (x) = 1 - x; h (x) = x + 1$ ; then show that $h \circ (g \circ f) = (h \circ g) \circ f$ .

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Solution

Given that:
$f, g, h$ are functions, $f (x) = x; g (x) = 1 - x; h (x) x + 1;$

To prove:
$h \circ (g \circ f) = (h \circ g) \circ f$

Solution:
L.H.S. $= h \circ (g \circ f)$
$g \circ f = 1 - x$
$h \circ (g \circ f) = h (1 - x)$
$h \circ (g \circ f) = 1 - x + 1$
$h \circ (g \circ f) = 2 - x$

R.H.S. $= (h \circ g) \circ f$
$(h \circ g) = 1 - x + 1 = 2 - x$
$(h \circ g) \circ f = 2 - x$
Therefore, L.H.S $=$ R.H.S

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Let f,g,h are functions defined by f(x)=x;g(x)=1−x;h(x)=x+1; then show that h∘(g∘f)=(h∘g)∘f.

Given that:f,g,h are functions, f(x)=x;g(x)=1−x;h(x)x+1;To prove: h∘(g∘f)=(h∘g)∘fSolution:L.H.S. =h∘(g∘f)g∘f=1−xh∘(g∘f)=h(1−x) h∘(g∘f)=1−x+1h∘(g∘f)=2−xR.H.S. =(h∘g)∘f(h∘g)=1−x+1=2−x(h∘g)∘f=2−xTherefore, L.H.S = R.H.S

Let $f, g, h$ are functions defined by $f (x) = x; g (x) = 1 - x; h (x) = x + 1$ ; then show that $h \circ (g \circ f) = (h \circ g) \circ f$ .