Question

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values as the case may be:
 $ g\left( x \right)=\dfrac{x}{2}+\dfrac{2}{x} $ , x>0

Answer 1

Hint: In this question, we need to find the local maxima or local minima and local maximum value or local minimum value for the given function g(x). For this, we will use the second derivative test. First we will find first derivative of g(x) i.e. g'(x) then we will put g'(x) = 0 and find value of x. Using this values of x in g''(x) (second derivative of g(x)), we will find value of g''(x). If g''(x)>0 for some value x, then that value of x will be local minima, if g''(x)<0 for any value of x, then that value of x will be local maxima. Putting the value of x in f(x) will give us the local maximum value or local minimum value.

Complete step by step answer:
Here we are given the function as,
 $ g\left( x \right)=\dfrac{x}{2}+\dfrac{2}{x} $ , x>0.
We know $ \dfrac{1}{x}={{x}^{-1}} $ so we get $ g\left( x \right)=\dfrac{x}{2}+2{{x}^{-1}} $ , x > 0
Let us first take derivative with respect to x on both sides, we get,
 $ g'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{x}{2} \right)+\dfrac{d}{dx}\left( 2{{x}^{-1}} \right) $ ,x>0.
Derivative of $ {{x}^{n}} $ is given as $ \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} $ and $ \dfrac{d}{dx}x=1 $ so we get,
 $ g'\left( x \right)=\dfrac{1}{2}+\left( -1 \right)2{{x}^{-1-1}}\Rightarrow g'\left( x \right)=\dfrac{1}{2}-2{{x}^{-2}}\Rightarrow g'\left( x \right)=\dfrac{1}{2}-\dfrac{2}{{{x}^{2}}} $ .
Now let us put g'(x) = 0 to find the value of x, we get,
 $ g'\left( x \right)=\dfrac{1}{2}-\dfrac{2}{{{x}^{2}}}=0\Rightarrow \dfrac{1}{2}=\dfrac{2}{{{x}^{2}}} $ .
Cross multiplying we get,
 $ {{x}^{2}}=4 $ .
Taking square root on both sides we get,
 $ x=\pm 2 $ .
Since x is given as greater than 0, so we get the value of x as 2 only.
Now let us find second derivative of g(x).
g'(x) is found as $ g'\left( x \right)=\dfrac{1}{2}-\dfrac{2}{{{x}^{2}}} $ .
Taking derivative with respect to x on both sides we get,
 $ g''\left( x \right)=\dfrac{d}{dx}\left( \dfrac{1}{2} \right)+\dfrac{d}{dx}\left( 2{{x}^{-2}} \right) $ .
We know that $ \dfrac{d}{dx}\left( a \right)=0 $ we get,
 $ g''\left( x \right)=0-2\left( -2 \right){{x}^{-2-1}}\Rightarrow g''\left( x \right)=4{{x}^{-3}}\Rightarrow g''\left( x \right)=\dfrac{4}{{{x}^{3}}} $ .
Putting x = 2 we get,
 $ g''\left( 2 \right)=\dfrac{4}{8}=\dfrac{1}{2} $ which is greater than 0.
Therefore 2 is the point of local minima by second derivative test.
Now let us find local minimum value. Putting x = 2 in g(x) we get, $ g\left( 2 \right)=\dfrac{2}{2}+\dfrac{2}{2}=1+1=2 $ .
Hence local minimum value is 2.

Note:
 Students should know the formula of derivatives for solving these sums. Take care of signs while finding a derivative. Do not take a value of x which is less than 0 because we are given the function as g(x) with x>0. Local minima are the value of x and the local minimum value is the value of g(x) when we put x as local minima.