Find the value of f(-1)+f(2)-f(4), where f(x)={[sqrt(2x-4)," for "x >= 4],[x," for "0 <= x < 4],[-2," for "x < 0]:}

Evaluate f(-1): Evaluate f(-1) using the definition of the function for x = 4. Since x = 4 is greater than or equal to 4, we use the first piece of the function f(x) = sqrt(2x - 4). f(4) = sqrt(2*4 - 4) = sqrt(8 - 4) = sqrt(4) = 2Combine results: Combine the results from steps 1, 2, and 3 to find the value of f(-1) + f(2) - f(4). f(-1) + f(2) - f(4) = (-2) + (2) - (2) = -2

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Find derivatives of logarithmic functions

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Q. Find the value of

f(-1)+f(2)-f(4)

, where

\newline

f(x)=\left\{\begin{array}{lll} \sqrt{2 x-4} & \text { for } x \geq 4 \\ x & \text { for } 0 \leq x<4 \\ -2 & \text { for } x<0 \end{array}\right.

Evaluate $f(-1)$ : Evaluate $f(-1)$ using the definition of the function for x < 0. $\newline$ Since $x = -1$ is less than $0$ , we use the third piece of the function $f(x) = -2$ . $\newline$ $f(-1) = -2$
Evaluate $f(2)$ : Evaluate $f(2)$ using the definition of the function for 0 \leq x < 4. $\newline$ Since $x = 2$ is between $0$ and $4$ , we use the second piece of the function $f(x) = x$ . $\newline$ $f(2) = 2$
Evaluate $f(4)$ : Evaluate $f(4)$ using the definition of the function for $x \geq 4$ . Since $x = 4$ is greater than or equal to $4$ , we use the first piece of the function $f(x) = \sqrt{2x - 4}$ . $f(4) = \sqrt{2\cdot4 - 4} = \sqrt{8 - 4} = \sqrt{4} = 2$
Combine results: Combine the results from steps $1$ , $2$ , and $3$ to find the value of $f(-1) + f(2) - f(4)$ . $f(-1) + f(2) - f(4) = (-2) + (2) - (2) = -2$