# What are the extremes of f(x)=ln(2x^2-20x+53)?

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We have to find the extremes of f(x) = ln(2x^2-20x+53).

At the extremes the value of the first derivative is 0.

f(x) = ln(2x^2-20x+53)

f'(x) = 4x - 20 / (2x^2 - 20x + 53)

4x - 20 / (2x^2 - 20x + 53) = 0

=> 4x - 20 = 0

=> x = 5

At x = 5 , f(x) = ln( 50 - 100 + 53)

=> ln 3

**The extreme of f(x) is (5 , ln 3)**

The local extremes of a function are the maximum or minimum points and they could be found in the critical points. Critical points are the roots of the 1st derivative of the function.

We'll differentiate the function using the chain rule.

f'(x) = (2x^2-20x+53)'/(2x^2-20x+53)

f'(x) = (4x - 20)/(2x^2-20x+53)

We'll solve the equation f'(x) = 0.

(4x - 20)/(2x^2-20x+53) = 0

This equation is cancelling when the numerator is cancelling.

4x - 20 = 0

4x = 20

x = 5

The critical point of the function is x = 5

For x<5 => f'(x)<0 => f(x) is decreasing.

For x>5, f'(x)>0 => f(x) is increasing.

Then x = 5 is a minimum point of the function.

We'll substitute x by 5 in the expression of f(x).

f(5) = ln 3

**The extreme point of the function is a minimum point and it has the coordinates (5 ; ln 3).**