How do you evaluate the function with the given values of $x : d f (x) = (\frac{1}{2}) [x + 1]; ddd x = 0; x = - 1$ $x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1$ ?

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How do you evaluate the function with the given values of $x : d f (x) = (\frac{1}{2}) [x + 1]; ddd x = 0; x = - 1$ $x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1$ ?

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Answer 1 · 2018-06-22T15:13:13

$f (0) \to y = \frac{1}{2}$ $f(0)->y=1/2$
$f (- 1) \to y = 0$ $f(-1)->y=0$

Explanation:

Suppose $x = 0$ $x=0$ then wherever there is an $x$ $x$ write 0

$f (x) = \frac{1}{2} (x + 1) ddd \to ddd f (0) = \frac{1}{2} (0 + 1) = \frac{1}{2}$ $f(x)=1/2(x+1) color(white)("ddd")-> color(white)("ddd")f(0)=1/2(0+1)=1/2$

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Suppose $x = - 1$ $x=-1$ then wherever there is an $x$ $x$ write -1

$f (x) = \frac{1}{2} (x + 1) ddd \to ddd f (0) = \frac{1}{2} [\frac{2}{2} (- 1) + 1 d] = 0$ $f(x)=1/2(x+1) color(white)("ddd")-> color(white)("ddd")f(0)=1/2[color(white)(2/2)(-1)+1color(white)("d")]=0$

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How do you evaluate the function with the given values of $x : d f (x) = (\frac{1}{2}) [x + 1]; ddd x = 0; x = - 1$ $x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1$ ?

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How do you evaluate the function with the given values of x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1x:df(x)=(12)[x+1];dddx=0;x=−1x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1?

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How do you evaluate the function with the given values of $x : d f (x) = (\frac{1}{2}) [x + 1]; ddd x = 0; x = - 1$ $x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1$ ?