Question

the distance a wave runner is from a dock on a shoreline is represented by the function ( f(d) = sqrt{49 + d^2} ), where ( d ) is the distance the wave runner has traveled since noon. the distance the wave runner has traveled is represented by the function ( g(t) = 6t ), where ( t ) is the time in hours the wave runner has traveled. what is ( (f circ g)(t) ) and what does it mean? what is the domain of ( (f circ g)(t) )? (1 point) ( (f circ g)(t) = sqrt{49 + 36t^2} ) represents the distance the wave runner is from the dock after ( t ) hours. the domain is ( t geq 0 ). ( (f circ g)(t) = 6sqrt{49 + t^2} ) represents the distance the wave runner travels. the domain is ( (-infty, infty) ). ( (f circ g)(t) = sqrt{49 + 36t^2} ) represents the distance the wave runner is from the dock after ( t ) hours. the domain is ( (-infty, infty) ). ( (f circ g)(t) = 6sqrt{49 + t^2} ) represents the distance the wave runner travels. the domain is ( t geq 0 ).

Explanation:

Step1: Find the composition \((f \circ g)(t)\)

To find the composition of two functions \(f(d)\) and \(g(t)\), we substitute \(g(t)\) into \(f(d)\). Given \(f(d)=\sqrt{49 + d^{2}}\) and \(g(t) = 6t\), we replace \(d\) in \(f(d)\) with \(g(t)=6t\). So, \((f\circ g)(t)=f(g(t))=f(6t)=\sqrt{49+(6t)^{2}}\). Simplify \((6t)^{2}\) which is \(36t^{2}\), so \((f\circ g)(t)=\sqrt{49 + 36t^{2}}\).

Step2: Determine the domain of \((f \circ g)(t)\)

The domain of a function is the set of all possible input values (in this case, values of \(t\)) for which the function is defined. For the square - root function \(y=\sqrt{u}\), the expression inside the square root (\(u\)) must be non - negative (i.e., \(u\geq0\)). In our function \((f\circ g)(t)=\sqrt{49 + 36t^{2}}\), the expression inside the square root is \(49+36t^{2}\). Since \(t^{2}\geq0\) for all real numbers \(t\) (because the square of any real number is non - negative), then \(36t^{2}\geq0\) for all real numbers \(t\). And \(49>0\), so \(49 + 36t^{2}>0\) for all real numbers \(t\). However, in the context of the problem, \(t\) represents time in hours since noon. Time cannot be negative (we can't have a negative number of hours since noon), so \(t\geq0\).

Now, let's analyze what \((f\circ g)(t)\) represents. \(g(t) = 6t\) represents the distance the wave runner has traveled (since speed is constant, distance = speed×time, and if we assume the speed is 6 units per hour, then distance traveled in \(t\) hours is \(6t\)). Then \(f(d)=\sqrt{49 + d^{2}}\) represents the distance of the wave runner from the dock, where \(d\) is the distance traveled. So, \((f\circ g)(t)\) represents the distance the wave runner is from the dock after \(t\) hours.

Looking at the options:

• Option 1: \((f\circ g)(t)=\sqrt{49 + 36t^{2}}\) represents the distance the wave runner is from the dock after \(t\) hours. The domain is \(t\geq0\). This matches our analysis.
• Option 2: The composition is wrong (it should be \(\sqrt{49 + 36t^{2}}\) not \(6\sqrt{49 + t^{2}}\)) and the domain description is incorrect.
• Option 3: The composition is wrong (it should be \(\sqrt{49 + 36t^{2}}\) not \(\sqrt{49 + 36t^{2}}\) with the wrong interpretation) and the domain \((-\infty,\infty)\) is incorrect in the context of time (time can't be negative).
• Option 4: The composition is wrong (it should be \(\sqrt{49 + 36t^{2}}\) not \(6\sqrt{49 + t^{2}}\)) and the domain description is incorrect.
• Option 5: The composition is wrong (it should be \(\sqrt{49 + 36t^{2}}\) not \(6\sqrt{49 + t^{2}}\)) and the domain description is incorrect.

Answer:

\((f\circ g)(t)=\sqrt{49 + 36t^{2}}\) represents the distance the wave runner is from the dock after \(t\) hours. The domain is \(t\geq0\) (the first option in the list of options provided in the problem).