What are the extrema of #f(x)=3x^2 - 12x + 13# on #[-oo,oo]#?

Answer 1

#f(x)# has a minimum at #x=2#

Before proceeding, note that this is an upwards facing parabola, meaning we can know without further calculation that it will have no maxima, and a single minimum at its vertex. Completing the square would show us that #f(x) = 3(x-2)^2+1#, giving the vertex, and thus the sole minimum, at #x = 2#. Let's see how this would be done with calculus, though.
Any extrema will occur either at a critical point or at an endpoint of the given interval. As our given interval of #(-oo,oo)# is open, we can ignore the possibility of endpoints, and so we will first identify the critical points of the function, that is, the point at which the derivative of the function is #0# or does not exist.
#f'(x) = d/dx (3x^2-12x+13) =6x-12#
Setting this equal to #0#, we find a critical point at #x=2#
#6x-12 = 0 => x = 12/6 = 2#
Now, we can either test to see whether it is an extremum (and what type) by checking some values of #f# around that point, or by using the second derivative test. Let's use the latter.
#(d^2x)/(dx^2) = d/dx(6x-12) =6#
As #f''(2) = 6 > 0#, the second derivative test tells us that #f(x)# has a local minimum at #x=2#
Thus, using #f'(x)# and #f''(x)#, we find that #f(x)# has a minimum at #x=2#, matching the result we found using algebra.
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Answer 2

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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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.

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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