Understanding Function Composition (gof and fog)

Function composition is the application of one function to the results of another. Given two functions, f and g, the composition \(g \circ f\), pronounced "g of f", is defined by \((g \circ f)(x) = g(f(x))\). Similarly, the composition \(f \circ g\), pronounced "f of g", is defined by \((f \circ g)(x) = f(g(x))\).

In this question, we are given two functions:

• \(f(x) = \cos x\)
• \(g(x) = 3x^2\)

Both functions map from the set of real numbers (R) to the set of real numbers (R).

Calculating gof(x)

To find \(gof(x)\), we substitute \(f(x)\) into \(g(x)\). The formula is \((g \circ f)(x) = g(f(x))\).

Substitute \(f(x) = \cos x\) into the expression for \(g(x)\):

\[ g(f(x)) = g(\cos x) \]

Since \(g(x) = 3x^2\), we replace \(x\) with \(\cos x\) in the expression for \(g(x)\):

\[ g(\cos x) = 3(\cos x)^2 = 3\cos^2 x \]

So, \(gof(x) = 3\cos^2 x\).

Calculating fog(x)

To find \(fog(x)\), we substitute \(g(x)\) into \(f(x)\). The formula is \((f \circ g)(x) = f(g(x))\).

Substitute \(g(x) = 3x^2\) into the expression for \(f(x)\):

\[ f(g(x)) = f(3x^2) \]

Since \(f(x) = \cos x\), we replace \(x\) with \(3x^2\) in the expression for \(f(x)\):

\[ f(3x^2) = \cos(3x^2) \]

So, \(fog(x) = \cos(3x^2)\).

Comparing gof(x) and fog(x)

We found that:

• \(gof(x) = 3\cos^2 x\)
• \(fog(x) = \cos(3x^2)\)

Let's check if these two expressions are equal for all values of \(x \in R\). Consider a simple value, for example, \(x = \pi\).

• \(gof(\pi) = 3\cos^2(\pi) = 3(-1)^2 = 3 \times 1 = 3\)
• \(fog(\pi) = \cos(3\pi^2)\)

Since \(3 \ne \cos(3\pi^2)\), we can see that \(gof(x)\) is not equal to \(fog(x)\) for all values of \(x\).

Therefore, \(gof \ne fog\).

Existence of gof and fog

The domain of \(f(x) = \cos x\) is R, and its range is \([-1, 1]\). The domain of \(g(x) = 3x^2\) is R, and its range is \([0, \infty)\).

For \(gof\) to exist, the range of \(f\) must be a subset of or equal to the domain of \(g\). The range of \(f\) is \([-1, 1]\), which is a subset of the domain of \(g\) (R). Thus, \(gof\) exists.

For \(fog\) to exist, the range of \(g\) must be a subset of or equal to the domain of \(f\). The range of \(g\) is \([0, \infty)\), which is a subset of the domain of \(f\) (R). Thus, \(fog\) exists.

Both \(gof\) and \(fog\) exist and map from R to R.

Conclusion

Based on our calculations, \(gof(x) = 3\cos^2 x\) and \(fog(x) = \cos(3x^2)\). These two functions are not equal. Both compositions exist.

Comparing this with the given options:

• Option 1: \(gof \ne fog\) - This matches our conclusion.
• Option 2: \(gof\) and \(fog\) both do not exist - This is incorrect as both exist.
• Option 3: \(gof = fog\) - This is incorrect.
• Option 4: Can't be determined - This is incorrect as we were able to determine the relationship.

Therefore, the correct statement is \(gof \ne fog\).