Solveeit Logo

Question

Question: How do you find the local extrema for\[f(x) = 5x - {x^2}\]?...

How do you find the local extrema forf(x)=5xx2f(x) = 5x - {x^2}?

Explanation

Solution

In the given question we have to find the extrema of the function f(x)=5xx2f(x) = 5x - {x^2}. Here extrema means the point of maximum for the given function. Now to find the point of maximum you can use the derivation method where you just have to derive the given function and keep it equal to 00. You will get a value of xx. Now find the value of yy by putting the value of xx you got in the previous step in the given function. These coordinates you got are the coordinate of point of maximum or local extrema.

Complete step by step solution:
In the given question we have to find the local extrema point, or you can say the point of maximum of the given function.
For that first of all derive the given function i.e.
f(x)=5xx2f(x)=52xf(x) = 5x - {x^2} \Rightarrow f'(x) = 5 - 2x
Now put f(x)=0f'(x) = 0in order to find the x coordinate i.e.

52x=0 2x=5 x=52  5 - 2x = 0 \\\ \Rightarrow 2x = 5 \\\ \Rightarrow x = \dfrac{5}{2} \\\

Now let y=5xx2y = 5x - {x^2} and put x=52x = \dfrac{5}{2}i.e.

y=5(52)(52)2 y=252254  y = 5(\dfrac{5}{2}) - {(\dfrac{5}{2})^2} \\\ y = \dfrac{{25}}{2} - \dfrac{{25}}{4} \\\

Further simplifying we get:

y=50254 y=254  y = \dfrac{{50 - 25}}{4} \\\ y = \dfrac{{25}}{4} \\\

Now we have x=52x = \dfrac{5}{2} and y=254y = \dfrac{{25}}{4} as the coordinate of the point of maximum or local extrema of the given function.

Hence, (52,254)(\dfrac{5}{2},\dfrac{{25}}{4}) is the required local extrema.

Note: Here first of all you should know the differentiations of simple functions, learn all the derivatives of basic functions. Now you have to be careful while doing the differentiation because most of the students make mistakes while differentiating the terms. This is one of the shortest methods to find the point of maximum of any function. You can verify this point by plotting a graph of the given function which you should because it would help you to learn how to plot graphs of different functions.