(Solved): Following function has a maximum point at f(x)=x242x+5 Your answer: x=4x=1x=4x=2 ...

Expert Answer

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To find the maximum point of the function f(x)=x2?42x+5?, we need to determine the values of x at which the function reaches its maximum.
To find the maximum point, we can take the derivative of the function and set it equal to zero. Let's differentiate the function f(x) with respect to x:
f?(x)=dxd?(x2?42x+5?)
Using the quotient rule, we can differentiate the function:
f?(x)=(x2?4)2(2)(x2?4)?(2x+5)(2x)?
Simplifying the numerator:
f?(x)=(x2?4)22x2?8?4x2?10x?
Combining like terms:
f?(x)=(x2?4)2?2x2?10x?8?
Setting f?(x) equal to zero to find critical points:
?2x2?10x?8=0
Solving this quadratic equation, we find two critical points:
x=?4andx=?1
To verify that these are the maximum points, we can check the second derivative. Let's find f??(x), the derivative of f?(x):
f??(x)=dxd?((x2?4)2?2x2?10x?8?)
Differentiating using the quotient rule:
f??(x)=(x2?4)4(?2)(2)(x2?4)2?(?2x2?10x?8)(2)(2x)(x2?4)?
Simplifying the numerator:
f??(x)=(x2?4)4?4(x2?4)2?4x(x2?4)(?2x2?10x?8)?