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Section 3-4 : The Definition of a Function

9. Given \(h\left( t \right) = {t^2} + 6\) determine each of the following.

  1. \(h\left( 0 \right)\)
  2. \(h\left( { - 2} \right)\)
  3. \(h\left( 2 \right)\)
  4. \(h\left( {\sqrt x } \right)\)
  5. \(h\left( {3 - t} \right)\)

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a \(h\left( 0 \right)\) Show Solution

Remember that for function evaluation all we need to do is replace all the \(t\)’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

\[h\left( 0 \right) = {\left( 0 \right)^2} + 6 = \require{bbox} \bbox[2pt,border:1px solid black]{6}\]

b \(h\left( { - 2} \right)\) Show Solution

Remember that for function evaluation all we need to do is replace all the \(t\)’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

\[h\left( { - 2} \right) = {\left( { - 2} \right)^2} + 6 = 4 + 6 = \require{bbox} \bbox[2pt,border:1px solid black]{{10}}\]

c \(h\left( 2 \right)\) Show Solution

Remember that for function evaluation all we need to do is replace all the \(t\)’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

\[h\left( 2 \right) = {\left( 2 \right)^2} + 6 = 4 + 6 = \require{bbox} \bbox[2pt,border:1px solid black]{{10}}\]

d \(h\left( {\sqrt x } \right)\) Show Solution

Remember that for function evaluation all we need to do is replace all the \(t\)’s (only one in this case) with whatever is in the parenthesis. It doesn’t matter if the stuff in the parenthesis is not a number the evaluation works exactly the same as if it was a number.

Doing the evaluation for this part gives,

\[h\left( {\sqrt x } \right) = {\left( {\sqrt x } \right)^2} + 6 = \require{bbox} \bbox[2pt,border:1px solid black]{{x + 6}}\]

e \(h\left( {3 - t} \right)\) Show Solution

Remember that for function evaluation all we need to do is replace all the \(t\)’s (only one in this case) with whatever is in the parenthesis. It doesn’t matter if the stuff in the parenthesis is not a number the evaluation works exactly the same as if it was a number. Do not get excited about the fact that the stuff in the parenthesis also involves an \(t\). Again, it does not matter, the evaluation still works the same way.

Doing the evaluation for this part gives,

\[h\left( {3 - t} \right) = {\left( {3 - t} \right)^2} + 6 = 9 - 6t + {t^2} + 6 = \require{bbox} \bbox[2pt,border:1px solid black]{{{t^2} - 6t + 15}}\]