Question

Consider the function f(X)-sinx and g(X)=x^3.find fog(X).show that the function fog(X)is a continuous function

Answer 1

Answer:

Given,

f(x) = sinx

g(x) = x³

∴fog(x) = f[g(x)] = f(x³) = sin(x³)

Now, for all values of x, -1≤ sin(x³) ≤1

∴ fog(x) = sin(x³) is continuous for all real values of x.

Answer 2

$sinx^3$

Step-by-step explanation:

f(x) = $sinx$

g(x) = $x^3$

fog is nothing but the value of the the function g(x) as a function to f(x).

fog(x) = f(g(x))

     = f($x^3$)

     = $sinx^3$

Now, to show that fog(x) is continuous, we must prove 3 things:

1. f(x) exists for all real values of x or f(c) exists.
2. $x\\lim_{x->c} f(x)$ must exist
3. f(c) = $x\\lim_{x->c} f(x)$

For all real values of x, -1 ≤ $sinx^3$ ≤ 1

Thus, we can see that all three conditions are satisfied as fog(x) is a trigonometric function. Trigonometric functions are continuous at all real values of the function, x.

Therefore, fog(x) is a continuous function.