Question

Show that the function is always increasing


`\sf{f(x) = log(1 + x) - (2x)/(x + 2)f(x)=log(1+x)−x+22x​}\ \textless \ br /\ \textgreater \ ​`
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Answer 1

Explanation:

`\large\underline{\sf{Solution-}}`

☼︎~Given function is

`\rm \: f(x) = log(1 + x) - (2x)/(x + 2)`

~Let first define the domain of f(x).

• Now, log(1 + x) is defined when x + 1 > 0

☼︎~Now, Consider

`\sf \: f(x) = log(1 + x) - (2x)/(x + 2)`

☼︎~On differentiating both sides

w. r. t. x, we get

`\rm {\: \longmapsto(d)/(dx)f(x) =(d)/(dx)\bigg[ log(1 + x) - (2x)/(x + 2)}`

☼︎~We know,

`\boxed{\sf{ \longmapsto(d)/(dx)logx \: = \: (1)/(x)}}`

❀And

`\boxed{\sf{\longmapsto (d)/(dx) (u)/(v) = \frac{v(d)/(dx)u \: - \: u(d)/(dx)v}{ {v}^(2) } \: }}`

~So, using these results, we get

`\sf {\:\longmapsto f'(x) = (1)/(x + 1) - \frac{(x + 2)(d)/(dx)2x - 2x(d)/(dx)(x + 2)}{ {(x + 2)}^(2) }}`

☼︎~We know,

`\boxed{\rm{\longmapsto (d)/(dx)x = \sf \pink{ 1 \:} }}`

And-

`\begin{gathered}\boxed{\sf{\longmapsto(d)/(dx)k = 0 \: }} \\ \end{gathered}`

☼︎~So, using this result, we get

`\rm \: f'(x) = (1)/(x + 1) - \frac{(x + 2)2 - 2x(1 + 0)}{ {(x + 2)}^(2) }f′(x)=x+11−(x+2)2(x+2)2−2x(1+0)</strong></p><p><strong>[tex]\rm \: f'(x) = (1)/(x + 1) - \frac{(x + 2)2 - 2x(1 + 0)}{ {(x + 2)}^(2) }f′(x)=x+11−(x+2)2(x+2)2−2x(1+0)\rm \: f'(x) = (1)/(x + 1) - \frac{2x + 4 - 2x}{ {(x + 2)}^(2) }`

`\rm \: f'(x) = (1)/(x + 1) - \frac{4 }{ {(x + 2)}^(2) }f′(x)=x+11−(x+2)24`

`\rm \: f'(x) = \frac{ {(x + 2)}^(2) - 4(x + 1)}{(x + 1) {(x + 2)}^(2) }`

`\rm \: f'(x) = \frac{ {x}^(2) + 4 + 4x - 4x - 4}{(x + 1) {(x + 2)}^(2) }`

`\rm \: f'(x) = \frac{ {x}^(2) }{(x + 1) {(x + 2)}^(2) }`

☼︎~Now, as

`\longmapsto\rm {\: {x}^(2) \geqslant 0x2⩾0}`

`</strong><strong>\longmapsto\sf\</strong><strong>:</strong><strong>{(x+2)}^(2) > 0`

`\longmapsto\sf\: x + 1 > 0x+1>0`

☼︎~So,

`\sf\longmapsto \:\rm \: f'(x) = \frac{ {x}^(2) }{(x + 1) {(x + 2)}^(2) } \geqslant 0`

☼︎~Therefore,

• f'(x) is always increasing when x > - 1

Additional Information:-

`\boxed{\begin{array}c \bf f(x) & \tt (d)/(dx)f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^(2)x \\ \\ \sf cotx & \sf - {cosec}^(2)x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf √(x) & \sf (1)/(2 √(x) ) \\ \\ \sf logx & \sf (1)/(x)\\ \\ \sf {e}^(x) & \sf {e}^(x) \end{array}}`