If $f (x) = 7 | x - 2 |$ $f(x)=7|x-2|$ , find $f (0)$ $f(0)$ , $f (- 2)$ $f(-2)$ , $f (x + 1)$ $f(x+1)$ and $f (x^{2} + 2)$ $f(x^2+2)$ ?

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Answer 1 · 2016-10-11T08:36:15

$f (0) = 14$ $f(0)=14$ , $f (- 2) = 28$ $f(-2)=28$ , $f (x + 1) = 7 | x - 1 |$ $f(x+1)=7|x-1|$ and $f (x^{2} + 2) = 7 x^{2}$ $f(x^2+2)=7x^2$

As $f (x) = 7 | x - 2 |$ $f(x)=7|x-2|$

$f (0) = 7 | 0 - 2 | = 7 \times | - 2 | = 7 \times 2 = 14$ $f(0)=7|0-2|=7xx|-2|=7xx2=14$

$f (- 2) = 7 | - 2 - 2 | = 7 \times | - 4 | = 7 \times 4 = 28$ $f(-2)=7|-2-2|=7xx|-4|=7xx4=28$

$f (x + 1) = 7 | x + 1 - 2 | = 7 | x - 1 |$ $f(x+1)=7|x+1-2|=7|x-1|$

$f (x^{2} + 2) = 7 ∣ ∣ x^{2} + 2 - 2 ∣ ∣ = 7 ∣ ∣ x^{2} ∣ ∣ = 7 x^{2}$ $f(x^2+2)=7|x^2+2-2|=7|x^2|=7x^2$ *as $x^{2}$ $x^2$ is always positive.

If f(x)=7|x−2|f(x)=7|x-2|, find f(0)f(0), f(−2)f(-2), f(x+1)f(x+1) and f(x2+2)f(x^2+2)?