Given,
![\[f\text{ }\left( x \right)\text{ }=\text{ }2x\text{ }+\text{ }bed\text{ }1\text{ }x\text{ }+\text{ }log\text{ }\left[ \surd \left( 1\text{ }+\text{ }x2 \right)\text{ }\text{ }x \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-34094e09186d15a45bf8e7e68846a45a_l3.png "Rendered by QuickLaTeX.com")
differentiating the two sides w.r.t. x, we get
On figuring out both the sides, we get
![\[4x4\text{ }+\text{ }1\text{ }+\text{ }4x2\text{ }\ge \text{ }1\text{ }+\text{ }x2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a4c647554c7adb03948bb5d403dc0ebe_l3.png "Rendered by QuickLaTeX.com")
![\[4x4\text{ }+\text{ }4x2\text{ }\text{ }x2\text{ }\ge \text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0531ba2a6b7ee920619deea704eee326_l3.png "Rendered by QuickLaTeX.com")
![\[4x4\text{ }+\text{ }3x2\text{ }\ge \text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-300985c680fa1f73d2e415d6ff68f5ca_l3.png "Rendered by QuickLaTeX.com")
![\[x2\left( 4x2\text{ }+\text{ }3 \right)\text{ }\ge \text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-03dea81a5ee0acd7da8192042e56c034_l3.png "Rendered by QuickLaTeX.com")
The above is valid for any worth of x ∈ R.
In this manner, the given capacity is an expanding capacity over R.
Given,
differentiating the two sides w.r.t. x, we get
On figuring out both the sides, we get
The above is valid for any worth of x ∈ R.
In this manner, the given capacity is an expanding capacity over R.