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Addition of Negative Numbers

20. If f( 2x ...

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20. If f( 2x + 1/x) = x + 1/4x + 1 (x is not equal to 0) , the value of f(x) is (1) 4x (2) 1/4(2x + 1/x) (3) 1/4x (4) 4(2x + 1/x)

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x \neq 0

∣ ∣ ∣ ∣ \begin{matrix} x + 1 & 2 x + 1 & 3 x + 1 2 x & 4 x + 3 & 6 x + 3 4 x + 4 & 6 x + 4 & 8 x + 4 \end{matrix} ∣ ∣ ∣ ∣ = 0

x + 1

Solve the following equations for x:

(i) $7^{2 x + 3} = 1$

(ii) $2^{x + 1} = 4^{x - 3}$

(iii) $2^{5 x + 3} = 8^{x + 3}$

(iv) $4^{2 x} = \frac{1}{32}$

(v) $4^{x - 1} \times (0.5)^{3 - 2 x} = {(\frac{1}{8})}^{x}$

(vi) $2^{3 x - 7} = 256$

f (x) = \frac{4 x + x^{4}}{1 + 4 x^{3}}

g (x) = ln (\frac{1 + x}{1 - x})

, then what is the value of

f \cdot g (\frac{e - 1}{e + 1})

f (x) = {\begin{matrix} x^{2} - 4 x + 3, & x < 2 x - 4, & x \geq 2 \end{matrix}

g (x) = {\begin{matrix} x - 3, & x < 3 x^{2} + 2 x + 2, & x \geq 3 \end{matrix}

\frac{f}{g} \cdot (1)

f (x) = x^{2} + 1, x \leq 0

= 2 x - 1, 0 < x < 5

= 4 x + 3, x \geq 5

\frac{f (- 3) + f (2) + f (5)}{f (1)} =

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