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For the following piecewise functions, evaluate f(_3), f(_2), f(4) and f(2)

F(x)=2x+1, x>0
=3x_1, x<0

1 answer

It looks like there may be a minor confusion in the definition of the piecewise function \( F(x) \). I'll interpret the function as follows:

\[ F(x) = \begin{cases} 2x + 1 & \text{if } x > 0 \ 3x & \text{if } x < 0 \end{cases} \]

Now, we will evaluate \( F(-3) \), \( F(-2) \), \( F(4) \), and \( F(2) \).

  1. Evaluate \( F(-3) \):

    • Since \(-3 < 0\), we use the second piece. \[ F(-3) = 3(-3) = -9 \]

  2. Evaluate \( F(-2) \):

    • Since \(-2 < 0\), we again use the second piece. \[ F(-2) = 3(-2) = -6 \]

  3. Evaluate \( F(4) \):

    • Since \(4 > 0\), we use the first piece. \[ F(4) = 2(4) + 1 = 8 + 1 = 9 \]

  4. Evaluate \( F(2) \):

    • Since \(2 > 0\), we use the first piece. \[ F(2) = 2(2) + 1 = 4 + 1 = 5 \]

Now, summarizing the evaluations:

  • \( F(-3) = -9 \)
  • \( F(-2) = -6 \)
  • \( F(4) = 9 \)
  • \( F(2) = 5 \)