If f- R- R is defined by x-2 x- -1 x3 -1- x- 1 2-x x- 1 - find the product f-0-5 f-3-

For 1≤ x≤ 1 f( x ) = x ^ 3 ⇒ f( .5)= ( .5) ^ 3 = .125 For x≤ 1 ⇒ f( 3)= 3+2= 1 ⇒ f( .5)f( 3)=( .125)( 1)=.125 ...

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Real Valued Functions

If f: R→ R ...

Question

If $f : R \to R$ is defined by $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & x \leq - 1 x^{3} & - 1 \leq x \leq 1 2 - x & x \geq 1 \end{matrix}$ , find the product $f (- 0.5) f (- 3)$ .

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f : R \to R

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & (x \leq - 1) x^{2} & (- 1 < x < 1) 2 - x & (x \geq 1) \end{matrix}

f (- 1) + f (0) + f (1)

f : R \to R

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & x \leq - 1 x^{2} & - 1 < x < 1 2 - x & x \geq 1 \end{matrix}

f (- 1.75) + f (0.5) + f (1.5)

f : R \to

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 2 & (x \leq - 1) x^{2} & (- 1 \leq x \leq 1) 2 - x & (x \geq 1) \end{matrix}

f (- 1.75) + f (0.5) + f (1.5)

f : R \to R

f (x) = \frac{2 x + 1}{3}

f^{- 1} (x) =

f : R \to R

f (x) = 2 x + 3

f^{- 1} (4)

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