If ` f:R^+→R^+` and ` g:R^+→R^+` , defined as ` f(x)=x^2,g(x)= sqrtx` ​, then find ` gof` and ` fog` whether are they equivalent?

Explanation

(i) At x = 4, f(x) is continuous and `underset(x to 4)"lim"f(x)= 8`
(ii)` f(x) = 6" for " x in [-4, -2)" Thus "underset(x to -3)"lim" f (x) = 6`.
(iii) When x approaches 0 from its right-hand side, the value of f(x) approaches - 1.
Hence `underset(x to 0^(+))"lim" f(x) =- 1`.
(iv) When x approaches 0 from its left hand, the value of f(x) approaches - 2.
Hence `underset(x to 0^(-))"lim" f(x) =- 2`.
(v) Clearly ` underset(x to 0)"lim" f(x)" does not exist as "underset(x to 0^(+))"lim" f(x) ne underset(x to 0^(-))" lim" f(x)`.
(vi) f(-2) = 8, see the dot at point (-2, 8) in the graph.
(vii) When x approaches 2 from its left - hand side, the value of f(x) approaches 7.
Hence ` underset(x to 2^(-))"lim" f (x) = 7`.
(viii) When x approaches - 2 from its left-hand side, the value of f(x) is exactly 6 as `f(x) = 6" for " x in [-4, -2)`. Hence `underset(x to 2^(-))"lim" f(x) = 6 ` (exact).
(ix) Since y = f(x) is continuous at ` x = 1 , underset(x to 0)"lim" f(x) = f(1) = 5`.
(x) f (0) =- 1.5.
(xi) ` underset(x to 0^(+))"lim" f(x-2) ` = value of f(x) to which it is approaching when x is approaching - 2 from its right - hand side. From the graph , this value is 2. Hence ` underset(x to 0^(+))"lim" f (x - 2) = 2`.
(xii) ` underset(x to 1^(-))"lim" f(x-4)` = value of f (x) to which it is approaching when x is approaching - 3 from its left-hand side. Since ` f(x) = 6" for " x in [-4, -2), underset(x to 1^(-))"lim" f(x - 4) = 6`.