If a function `f: R->R` be defined by `f(x)={3x-2,\ x<0 1, x=0 4x+1, x >0` Find: `f(10 ,\ f(-1),\ f(0),\ f(2)dot`

Explanation

Given ,

If x > 0, `f (x) = 4x + 1 `

By substituting x = 1 we get

`f (1) = 4(1) + 1 = 4 + 1 = 5 `


If x < 0, `f(x) = 3x – 2`

By substituting x = –1 we get

` f (–1) = 3(–1) – 2 = –3 – 2 = –5`

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