For $f (t) = | t | + 1$ $f(t)=abs[t]+1$ ; how do you find f(-5),f(0) and f(-9)?

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Answer 1 · 2015-07-07T22:26:00

$f (- 5) = 6$ $f(-5) = 6$
$f (0) = 1$ $f(0) = 1$
$f (- 9) = 10$ $f(-9) = 10$

Given $f (t) = | t | + 1$ $f(t)=abs(t)+1$

For specific values of $t$ $t$ :
$f (- 5) = | - 5 | + 1 = 5 + 1 = 6$ $f(-5) = abs(-5)+1 = 5+1 = 6$

$f (0) = | 0 | + 1 = 0 + 1 = 1$ $f(0) = abs(0) + 1 = 0+1 = 1$

$f (- 9) = | - 9 | + 1 = 9 + 1 = 10$ $f(-9) = abs(-9)+1 = 9+1 =10$

For f(t)=abs[t]+1f(t)=|t|+1f(t)=abs[t]+1; how do you find f(-5),f(0) and f(-9)?