What are the extrema of #f(x)=3x^2 - 12x + 13# on #[-oo,oo]#?
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To find the extrema of ( f(x) = 3x^2 - 12x + 13 ) on the interval ( [-\infty, \infty] ), we first find the critical points by taking the derivative of the function and setting it equal to zero.
Taking the derivative of ( f(x) ) with respect to ( x ), we get:
( f'(x) = 6x - 12 )
Setting ( f'(x) ) equal to zero and solving for ( x ), we find the critical point:
( 6x - 12 = 0 )
( 6x = 12 )
( x = 2 )
Now, we need to determine whether this critical point corresponds to a maximum, minimum, or neither. We can do this by checking the second derivative of ( f(x) ) at ( x = 2 ).
Taking the second derivative of ( f(x) ), we get:
( f''(x) = 6 )
Since the second derivative is positive, the critical point ( x = 2 ) corresponds to a minimum.
Therefore, the function ( f(x) = 3x^2 - 12x + 13 ) has a minimum at ( x = 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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