A function f - -3- 7- R is defined as follows- fx - 4x2 -1- -3 - x - 2 3x - 2- 2- x - 4 2x - 3- 4 - x - 6-Find- f5 - f6

Given, f(x) = { #160;4x^2 1; amp; 3 ≤ x lt; 2 3x 2; #160; amp; 2≤ x ≤ 4 2x 3; #160; amp; 4 lt; x ≤ 6. Then, f(5) = 2 ...

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Question

A function $f : (- 3, 7) \to R$ is defined as follows:
$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 4 x^{2} - 1; & - 3 \leq x < 2 3 x - 2; & 2 \leq x \leq 4 2 x - 3; & 4 < x \leq 6 \end{matrix}$
Find: $f (5) + f (6)$

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f : [- 3, 7) \to R

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 4 x^{2} - 1; & - 3 \leq x < 2 3 x - 2; & 2 \leq x \leq 4 2 x - 3; & 4 < x < 7 \end{matrix}

f (5) + f (6)

f (1) - f (3)

f (- 2) - f (- 4)

\frac{f (3) + f (1)}{2 f (6) - f (1)}

f : [- 3, 7) \to R

f (x) =

⎛ ⎜ ⎝ \begin{matrix} 4 x^{2} - 1; - 3 \leq x < 2 3 x - 2; 2 \leq x \leq 4 2 x - 3; 4 < x < 7 \end{matrix}

\frac{f (3) + f (- 1)}{2 f (6) - f (1)}

f : [1, 6) \to R

f (x) =

⎧ ⎨ ⎩ \begin{matrix} 1 + x, 1 \leq x < 2 2 x - 1, 2 \leq x < 4 3 x^{2} - 10, 4 \leq x < 6 \end{matrix}

[1, 6) = {x \in R : 1 \leq x < 6}

f (5)

f (3)

f (1)

f (2) - f (4)

2 f (5) - 3 f (1)

f : [- 7, 6) \to R

f (x) = x^{2} + 2 x + 1

- 7 \leq x < - 5

x + 5

- 5 \leq x \leq 2

x - 1

2 < x < 6

(a) f (- 7) - f (- 3)

(b) \frac{4 f (- 3) + 2 f (4)}{f (- 6) - 3 f (1)}

f : [- 7, 6) \to R

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} + 2 x + 1; & - 7 \leq x < - 5 x + 5; & - 5 \leq x \leq 2 x - 1; & 2 < x < 6 \end{matrix}

2 f (- 4) + 3 f (2)

f (- 7) - f (- 3)

\frac{4 f (- 3) + 2 f (4)}{f (- 6) - 3 f (1)}

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