What are the points of inflection, if any, of #f(x)=1/(x^2-4x+5) #?

Answer 1

#x=(-(-12)+-sqrt((-12)^2-4(3)(11)))/(2(3))#

Rewrite the function as:

#(x^2-4x+5)^-1#

Then differentiate:

#-(x^2-4x+5)^-2(2x-4)#

Then differentiate again. But now, we're going to have to use chain rule:

#(-(x^2-4x+5)^-2)(2)+(2x-4)(-(-2)(x^2-4x+5)^-3(2x-4))#

Rewrite:

#2(2x+4)^2/(x^2-4x+5)^3 -2/(x^2-4x+5)^2#

Use common denominators and merge the fractions:

#(2(3x^2-12x+11))/(x^2-4x+5)^3#
Inflection points may happen when #f"(x)=0# So only worry about the top part of the fraction:
#2(3x^2-12x+11)=0#
#3x^2-12x+11=0#

Use quadratic formula:

#x=(-(-12)+-sqrt((-12)^2-4(3)(11)))/(2(3))#
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Answer 2

To find the points of inflection of ( f(x) = \frac{1}{x^2 - 4x + 5} ), we first find the second derivative and set it equal to zero to locate any inflection points.

The first derivative is ( f'(x) = -\frac{2x - 4}{(x^2 - 4x + 5)^2} ).

The second derivative is ( f''(x) = \frac{2(3x^2 - 8x + 13)}{(x^2 - 4x + 5)^3} ).

Setting ( f''(x) ) equal to zero and solving for ( x ) yields no real solutions.

Thus, there are no points of inflection for the function ( f(x) = \frac{1}{x^2 - 4x + 5} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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