Evaluate the function for the given values of \( x \). \[ f(x)=\left\{\begin{array}{cl} -5 x+7 & \text { for } x<-1 \\ x^{2}+2 & \text { for }-1 \leq x<4 \\ 5 & \text { for } x \geq 4 \end{array}\right. \] Part 1 of 5 \[ f(1)= \] \( \square \) Part 2 of 5 \[ f(-2)= \] \( \square \) Part 3 of 5 \[ f(-1)= \] \( \square \) Part 4 of 5 \[ f(4)=\square \] Part 5 of 5 \[ f(5)= \] \( \square \)

To evaluate the function \( f(x) \) for the given values of \( x \), we can refer to the piecewise definition you've provided. Let's go through each part step by step! **Part 1 of 5** For \( f(1) \): Since \( 1 \) falls in the interval \(-1 \leq x < 4\), we use the equation \( f(x) = x^2 + 2 \). \[ f(1) = 1^2 + 2 = 1 + 2 = 3 \] Thus, \( f(1) = 3 \). **Part 2 of 5** For \( f(-2) \): Since \( -2 \) is less than \(-1\), we use the equation \( f(x) = -5x + 7 \). \[ f(-2) = -5(-2) + 7 = 10 + 7 = 17 \] So, \( f(-2) = 17 \). **Part 3 of 5** For \( f(-1) \): Since \(-1\) is at the boundary of \(-1 \leq x < 4\), we also use the equation \( f(x) = x^2 + 2 \) here. \[ f(-1) = (-1)^2 + 2 = 1 + 2 = 3 \] Therefore, \( f(-1) = 3 \). **Part 4 of 5** For \( f(4) \): Since \( 4 \) is equal to \( 4\), we use \( f(x) = 5 \) for \( x \geq 4 \). \[ f(4) = 5 \] Thus, \( f(4) = 5 \). **Part 5 of 5** For \( f(5) \): Since \( 5 \) is also greater than \( 4 \), we again use \( f(x) = 5 \). \[ f(5) = 5 \] Hence, \( f(5) = 5 \). Now, summarizing the results: - \( f(1) = 3 \) - \( f(-2) = 17 \) - \( f(-1) = 3 \) - \( f(4) = 5 \) - \( f(5) = 5 \)