What are the extrema of #f(x)=-x^2 +5x -1 #?
relative max at
Use the 2nd derivative test to see if the critical number is a relative max. or relative min.:
Find the y-value of the maximum:
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To find the extrema of ( f(x) = -x^2 + 5x - 1 ), you first need to find its critical points.
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Take the derivative of the function: [ f'(x) = -2x + 5 ]
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Set the derivative equal to zero and solve for ( x ) to find critical points: [ -2x + 5 = 0 ] [ 2x = 5 ] [ x = \frac{5}{2} ]
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Test the critical point to determine if it is a maximum or minimum: [ f''(x) = -2 ] Since the second derivative is negative, the critical point ( x = \frac{5}{2} ) corresponds to a maximum.
Therefore, the maximum (extrema) of the function occurs at ( x = \frac{5}{2} ). To find the maximum value, substitute ( x = \frac{5}{2} ) back into the original function: [ f\left(\frac{5}{2}\right) = -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) - 1 ] [ f\left(\frac{5}{2}\right) = -\frac{25}{4} + \frac{25}{2} - 1 ] [ f\left(\frac{5}{2}\right) = \frac{25}{4} - 1 ] [ f\left(\frac{5}{2}\right) = \frac{25}{4} - \frac{4}{4} ] [ f\left(\frac{5}{2}\right) = \frac{21}{4} ]
So, the maximum value (extrema) of the function ( f(x) ) is ( \frac{21}{4} ) when ( x = \frac{5}{2} ).
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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.
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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