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### Section 3-4 : The Definition of a Function

7. Given $$f\left( x \right) = 3 - 2{x^2}$$ determine each of the following.

1. $$f\left( 0 \right)$$
2. $$f\left( 2 \right)$$
3. $$f\left( { - 4} \right)$$
4. $$f\left( {3t} \right)$$
5. $$f\left( {x + 2} \right)$$

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a $$f\left( 0 \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$x$$’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

$f\left( 0 \right) = 3 - 2{\left( 0 \right)^2} = \require{bbox} \bbox[2pt,border:1px solid black]{3}$

b $$f\left( 2 \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$x$$’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

$f\left( 2 \right) = 3 - 2{\left( 2 \right)^2} = 3 - 2\left( 4 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 5}}$

c $$f\left( { - 4} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$x$$’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

$f\left( { - 4} \right) = 3 - 2{\left( { - 4} \right)^2} = 3 - 2\left( {16} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 29}}$

d $$f\left( {3t} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$x$$’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,

$f\left( {3t} \right) = 3 - 2{\left( {3t} \right)^2} = 3 - 2\left( {9{t^2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{3 - 18{t^2}}}$

e $$f\left( {x + 2} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$x$$’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 $$x$$. Again, it does not matter, the evaluation still works the same way.

Doing the evaluation for this part gives,

$f\left( {x + 2} \right) = 3 - 2{\left( {x + 2} \right)^2} = 3 - 2\left( {{x^2} + 4x + 4} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 2{x^2} - 8x - 5}}$