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Suppose that the function h is defined, for all real numbers, as follows. h(x)={[(1)/(2)x-2," if "x <= -2],[-(x+1)^(2)," if "-2 < x <= 1],[-1," if "x > 1]:} Find h(-3),h(-1), and h(1). {:[h(-3)=],[h(-1)=],[h(1)=]:}

Suppose that the function h h is defined, for all real numbers, as follows.\newlineh(x)={12x−2amp; if x≤−2−(x+1)2amp; if −2<x1=""≤=""=""−1=""=""=""=""if=""=""x="">1 h(x)=\left\{\begin{array}{ll} \frac{1}{2} x-2 &amp; \text { if } x \leq-2 \\ -(x+1)^{2} &amp; \text { if }-2<x 1="" \leq="" \\="" -1="" &="" \text="" {="" if="" }="" x="">1 \end{array}\right. \newlineFind h(−3),h(−1) h(-3), h(-1) , and h(1) h(1) .\newlineh(−3)=h(−1)=h(1)= \begin{array}{l} h(-3)= \\ h(-1)= \\ h(1)= \end{array}

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Full solution

Q. Suppose that the function h h is defined, for all real numbers, as follows.\newlineh(x)={12x−2 if x≤−2−(x+1)2 if −2<x≤1−1 if x>1 h(x)=\left\{\begin{array}{ll} \frac{1}{2} x-2 & \text { if } x \leq-2 \\ -(x+1)^{2} & \text { if }-2<x \leq 1 \\ -1 & \text { if } x>1 \end{array}\right. \newlineFind h(−3),h(−1) h(-3), h(-1) , and h(1) h(1) .\newlineh(−3)=h(−1)=h(1)= \begin{array}{l} h(-3)= \\ h(-1)= \\ h(1)= \end{array}
  1. Evaluate h(−3)h(-3): We need to evaluate h(−3)h(-3). Since −3-3 is less than or equal to −2-2, we use the first part of the piecewise function: h(x)=12x−2h(x) = \frac{1}{2}x - 2.
    h(−3)=12(−3)−2h(-3) = \frac{1}{2}(-3) - 2
    h(−3)=−1.5−2h(-3) = -1.5 - 2
    h(−3)=−3.5h(-3) = -3.5
  2. Evaluate h(−1)h(-1): Next, we evaluate h(−1)h(-1). Since −1-1 is greater than −2-2 and less than or equal to 11, we use the second part of the piecewise function: h(x)=−(x+1)2h(x) = -(x + 1)^2.
    h(−1)=−(−1+1)2h(-1) = -(-1 + 1)^2
    h(−1)=−(0)2h(-1) = -(0)^2
    h(−1)=−0h(-1) = -0
    h(−1)=0h(-1) = 0
  3. Evaluate h(1)h(1): Finally, we evaluate h(1)h(1). Since 11 is less than or equal to 11, we use the second part of the piecewise function again: h(x)=−(x+1)2h(x) = -(x + 1)^2.
    h(1)=−(1+1)2h(1) = -(1 + 1)^2
    h(1)=−(2)2h(1) = -(2)^2
    h(1)=−4h(1) = -4

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