# What are the values and types of the critical points, if any, of #f(x)=x^2-5x+4#?

A Minimum point

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The critical points of ( f(x) = x^2 - 5x + 4 ) occur where the derivative is zero or undefined.

First, find the derivative: [ f'(x) = 2x - 5 ]

Next, set the derivative equal to zero and solve for ( x ) to find the critical points: [ 2x - 5 = 0 ] [ x = \frac{5}{2} ]

This critical point corresponds to a minimum since the second derivative is positive: [ f''(x) = 2 ]

Therefore, the critical point is ( \left(\frac{5}{2}, -\frac{9}{4}\right) ).

There are no other critical points, as the derivative is defined everywhere and does not equal zero for any other value of ( x ).

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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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