CBSE (Arts) Class 12CBSE
Share
Notifications

View all notifications

For the Function F(X) = X + 1 X (A) X = 1 is a Point of Maximum (B) X = − 1 is a Point of Minimum (C) Maximum Value > Minimum Value (D) Maximum Value< Minimum Value - CBSE (Arts) Class 12 - Mathematics

Login
Create free account


      Forgot password?

Question

For the function f(x) = \[x + \frac{1}{x}\]

  • x = 1 is a point of maximum

  • x = \[-\] 1 is a point of minimum

  • maximum value > minimum value

  • maximum value < minimum value

Solution

\[\text { maximum value < minimum value}\]

 

\[\text { Given:} f\left( x \right) = x + \frac{1}{x}\]

\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]

\[\text { For a local maxima or a local minima, we must have} \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]

\[ \Rightarrow x^2 - 1 = 0\]

\[ \Rightarrow x^2 = 1\]

\[ \Rightarrow x = \pm 1\]

\[\text { Now }, \]

\[f''\left( x \right) = \frac{2}{x^3}\]

\[ \Rightarrow f''\left( 1 \right) = \frac{2}{1} = 2 > 0\]

\[\text { So, x = 1 is a local minima.}\]

\[\text { Also }, \]

\[f''\left( - 1 \right) = - 2 < 0\]

\[\text {So, x = - 1 is a localmaxima }.\]

\[\text { The local minimum value is given by }\]

\[f\left( 1 \right) = 2\]

\[\text { The local maximum value is given by }\]

\[f\left( - 1 \right) = - 2\]

\[ \therefore \text { Maximum value < Minimum value }\]

  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

Solution For the Function F(X) = X + 1 X (A) X = 1 is a Point of Maximum (B) X = − 1 is a Point of Minimum (C) Maximum Value > Minimum Value (D) Maximum Value< Minimum Value Concept: Graph of Maxima and Minima.
S
View in app×