For $g(t)=t^3 +3$ ; how do you find g(1),g(-5),and g(0)?

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Answer 1 · 2015-08-01T22:55:22

A variable like $t$ is a kind of 'place holder' for any value or expression.

Put $1$ in place of $t$ to find: $g(1) = 4$

Similarly $g(-5) = -122$ and $g(0) = 3$

Put $color(red)(1)$ in place of $t$ to find:

$g(color(red)(1)) = (color(red)(1))^3+3 = 1+3 = 4$

Putting $color(red)(-5)$ in place of $t$ we find:

$g(color(red)(-5)) = (color(red)(-5))^3+3 = -125+3 = -122$

Putting $color(red)(0)$ in place of $t$ we find:

$g(color(red)(0)) = (color(red)(0))^3+3 = 0+3 = 3$

You can put expressions in place of $t$ too.

For example,

$g(color(red)(2x+1)) = (color(red)(2x+1))^3+3$

$= (8x^3+12x^2+6x+1)+3$

$= 8x^3+12x^2+6x+4$

For g(t)=t^3 +3g(t)=t^3 +3; how do you find g(1),g(-5),and g(0)?