Use \( f(x)=b x+j \) and \( g(x)=k x+p \) to find: \( \begin{array}{ll}\text { (a) } f \circ g & \text { (b) } g \circ f \\ \text { (c) the domain of } f \circ g \text { and of } g \circ f & \text { (d) the conditions for which } f \circ g=g \circ f \\ \text { (a) }(f \circ g)(x)=\square & \end{array} \)

Below is a step‐by‐step solution. --- ### (a) Find \( (f\circ g)(x) \) Recall that by definition, \[ (f\circ g)(x)=f(g(x)). \] Given \[ g(x)=kx+p, \] substitute this into \( f(x)=bx+j \) to obtain: \[ (f \circ g)(x) = f(kx+p) = b(kx+p) + j. \] Distribute \( b \): \[ (f \circ g)(x)=bkx+bp+j. \] Thus, the answer for part (a) is: \[ \boxed{(f \circ g)(x)=bkx+bp+j.} \] --- ### (b) Find \( (g\circ f)(x) \) Similarly, by definition, \[ (g\circ f)(x)=g(f(x)). \] Since \[ f(x)=bx+j, \] substitute this into \( g(x)=kx+p \) to get: \[ (g \circ f)(x) = g(bx+j)= k(bx+j)+p. \] Distribute \( k \): \[ (g \circ f)(x)= bkx+kj+p. \] Thus, the answer is: \[ \boxed{(g \circ f)(x)=bkx+kj+p.} \] --- ### (c) Determine the Domain of \( f\circ g \) and \( g\circ f \) Since both \( f(x)=bx+j \) and \( g(x)=kx+p \) are linear functions, they are defined for all real numbers. Therefore, their compositions are also defined for all real numbers. Thus, the domain for both \( f \circ g \) and \( g \circ f \) is: \[ \boxed{\text{All real numbers, } \mathbb{R}.} \] --- ### (d) Find the Conditions Under Which \( f\circ g = g\circ f \) For \( f\circ g \) and \( g\circ f \) to be equal for all \( x \), we must have: \[ bkx+bp+j = bkx+kj+p \quad \text{for all } x. \] Since the coefficients of \( x \) in both expressions are \( bk \) (which are already equal), equate the constant terms: \[ bp+j = kj+p. \] To simplify, subtract \( j+p \) from both sides: \[ bp - p = kj - j. \] Factor each side: \[ p(b-1) = j(k-1). \] So, the condition for which \( f\circ g = g\circ f \) is: \[ \boxed{p(b-1) = j(k-1).} \] --- ### Final Answer - **(a)** \(\displaystyle (f \circ g)(x)=bkx+bp+j\) - **(b)** \(\displaystyle (g \circ f)(x)=bkx+kj+p\) - **(c)** Domain is \(\displaystyle \mathbb{R}\) for both compositions. - **(d)** \( \displaystyle p(b-1)=j(k-1) \) This completes the step-by-step solution.