Question

Use the first derivative test to find local extrema of $ f\left( x \right) = {x^2} - 6x + 8 $ on $ R $ .

Answer 1

Hint: Local extrema of a function is the point at which a maximum or minimum value of the function in some interval containing the point is obtained. In order to find the local extrema of a function, we have to equate the first derivative of the function with zero as at that point, the slope of the function will be zero. So, here we will first differentiate the function $ f\left( x \right) $ to obtain $ f'\left( x \right) $ and then we will equate it with zero.

Complete step by step solution:
(i)
We are given,
 $ f\left( x \right) = {x^2} - 6x + 8 $
Firstly, we will differentiate $ f\left( x \right) $ with respect to $ x $ in order to obtain $ f'\left( x \right) $ . Therefore,
 $ f'\left( x \right) = 2x - 6 $
(ii)
Now, as we know that the first derivative $ f'\left( x \right) $ is also known as the slope of the function. We can say that when the slope of the function will be zero, we will obtain a maxima or a minima at that point. Therefore, equating $ f'\left( x \right) $ with zero, we will get:
 $
  f'\left( x \right) = 0 \\
  2x - 6 = 0 \\
  2x = 6 \\
  x = \dfrac{6}{2} \\
  x = 3 \;
  $
Since, we get $ x = 3 $ , it means that the slope of the function i.e., $ f'\left( x \right) $ attains the value $ 0 $ at $ x = 3 $ which implies that at $ x = 3 $ , the function has a local extrema i.e., either a maxima or a minima.
(iii)
In order to find whether it is a maxima or a minima, we will find the double derivative of the function. Therefore, we will differentiate the first derivative again to obtain the double derivative of the function. So, we will get:
 $ f''\left( x \right) = 2 $
Since, $ f''\left( x \right) $ at $ x = 3 $ is positive as it is a constant function, we can say that we have a local minima of $ f\left( x \right) $ at $ x = 3 $ .
(iv)
To find the value of this local minima, which we know lies on $ x = 3 $ , we will put the value of $ x $ as $ 3 $ in $ f\left( x \right) $ .
 $ f\left( 3 \right) = {\left( 3 \right)^2} - 6\left( 3 \right) + 8 $
Solving it further, we get:
 $
  f\left( 3 \right) = 9 - 18 + 8 \\
  f\left( 3 \right) = - 1 \;
  $
Hence, the local minima of the function $ f\left( x \right) = {x^2} - 6x + 8 $ lies on $ x = 3 $ and its value is $ f\left( 3 \right) = - 1 $ .
So, the correct answer is “-1”.

Note: We should always remember that the first derivative of the function is the slope of the function and the solutions of the first derivative will give us the points of local extrema of the function. In order to check whether the extrema is a local minima or a local extrema, we can find the double derivative which helps us to find the concavity of the function. If the value of the double derivative of the function is positive at the point of the local extrema, it is said to be a local minima whereas if the value of the double derivative comes out to be negative at the point of local extrema, it is said to be a local maxima. In this question, we got the double derivative as a constant function so we did not have to put the value of $ x $ as the point of local extrema, otherwise we would have put it as $ x = 3 $ .