Find the value

Question:

Let $f: R \rightarrow R: f(x)=x^{2}$ and $g: R \rightarrow R: g(x)=(x+1)$.

Show that $(g \circ f) \neq(f \circ g) .$

 

Solution:

To prove: $(g \circ f) \neq(f \circ g)$

Formula used: (i) g o f = g(f(x))

(ii) f o g = f(g(x))

Given: (i) $f: R \rightarrow R: f(x)=x^{2}$

(ii) $g: R \rightarrow R: g(x)=(x+1)$

Proof: We have,

$g \circ f=g(f(x))=g\left(x^{2}\right)=\left(x^{2}+1\right)$

fo $g=f(g(x))=g(x+1)=\left[(x+1)^{2}+1\right]=x^{2}+2 x+2$

From the above two equation we can say that (g o f) ≠ (f o g)

Hence Proved

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now