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What are the local extrema of #f(x)= -2x^2 + 9x#?

Answer 1

William Abdalla

We have a maxima at #x=0#

As #f(x)=-2x^2+9#, #f'(x)=-4x#

As #f'(x)=0# for #x=0#, hence we have a local extrema at #x=-9/4#

Further, #f''(x)=-4# and hence at #x=0#,

we have a maxima at #x=0#

graph{-2x^2+9 [-5, 5, -10, 10]}

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Answer 2

Luke Alvarez

To find the local extrema of ( f(x) = -2x^2 + 9x ), we first find the derivative of the function and set it equal to zero to locate critical points. Then, we determine the nature of these critical points by analyzing the sign of the second derivative.

Find the derivative: [ f'(x) = -4x + 9 ]
Set ( f'(x) ) equal to zero and solve for ( x ) to find critical points: [ -4x + 9 = 0 ] [ x = \frac{9}{4} ]
Determine the nature of the critical point using the second derivative test: [ f''(x) = -4 ]

Since the second derivative is constant, we can directly evaluate it at the critical point: [ f''\left(\frac{9}{4}\right) = -4 ]

Since the second derivative is negative, the function has a local maximum at ( x = \frac{9}{4} ).

So, the local maximum of ( f(x) = -2x^2 + 9x ) occurs at ( x = \frac{9}{4} ).

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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.