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What are the extrema of #f(x) = 3x^2 + 12x + 16#?

Answer 1

Ezra Adams

#x=-2# is a minimum.

#f'(x)=6x+12=0#

#x=-2#

#f''(-2)=6>0# #=># local minimum

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Answer 2

Christopher Agresta

To find the extrema of ( f(x) = 3x^2 + 12x + 16 ), we first take the derivative of the function and set it equal to zero to find the critical points. Then, we determine whether these critical points correspond to maximum or minimum values by using the second derivative test.

First derivative: ( f'(x) = 6x + 12 )

Setting ( f'(x) ) equal to zero: ( 6x + 12 = 0 ) ( 6x = -12 ) ( x = -2 )

Second derivative: ( f''(x) = 6 )

Since the second derivative is positive, the critical point ( x = -2 ) corresponds to a minimum value.

Therefore, the minimum (extrema) of the function ( f(x) = 3x^2 + 12x + 16 ) occurs at ( x = -2 ), and the minimum value is ( f(-2) = 8 ).

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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.