How do you find the critical points for #f(x)=x^(8)# and the local max and min?

Answer 1

Check below.

#f(x)=x^8#
#D_f=RR#
#f# is continuous in #RR#
#f'(x)=8x^7# and
#f'(x)=0 <=> x=0# #x_0=0# is a critical point. There is no other critical point because #f'# is defined in #RR#.
If #x<0# then #f'(x)<0# so #f# is decreasing in #(-oo,0]# If #x>0# then #f'(x)>0# so #f# is increasing in #[0,+oo)#
#0##<##x# #<=> f(x)>f(0)=0#
#x>=0# #<=> f(x)>=f(0)=0#
So for #x##in##RR#: #f(x)>=f(0)#
Therefore #f# has global minimum at #x_0=0# , #f(0)=0#
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Answer 2

To find the critical points for ( f(x) = x^8 ) and the local maxima and minima, we need to find where the derivative of the function equals zero.

( f'(x) = 8x^7 )

Setting ( f'(x) ) equal to zero:

( 8x^7 = 0 )

( x = 0 )

So, ( x = 0 ) is the only critical point.

To determine if ( x = 0 ) is a local maximum, local minimum, or neither, we can use the second derivative test.

( f''(x) = 56x^6 )

Evaluate ( f''(0) ):

( f''(0) = 56(0)^6 = 0 )

Since ( f''(0) = 0 ), the second derivative test is inconclusive.

Therefore, ( x = 0 ) is a critical point, but it is neither a local maximum nor a local minimum for the function ( f(x) = x^8 ).

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Answer 3

To find the critical points of ( f(x) = x^8 ) and determine the local maximum and minimum points, we first need to find the derivative of the function, set it equal to zero to find the critical points, and then use the second derivative test to determine if these critical points correspond to local maximum, minimum, or points of inflection.

Given the function ( f(x) = x^8 ), we find its derivative ( f'(x) ) using the power rule, which states that the derivative of ( x^n ) with respect to ( x ) is ( nx^{n-1} ):

[ f(x) = x^8 ] [ f'(x) = 8x^{8-1} ] [ f'(x) = 8x^7 ]

Now, to find the critical points, we set the derivative equal to zero and solve for ( x ):

[ 8x^7 = 0 ]

[ x = 0 ]

So, ( x = 0 ) is the only critical point.

To determine if this critical point corresponds to a local maximum, minimum, or a point of inflection, we use the second derivative test. The second derivative ( f''(x) ) is found by differentiating ( f'(x) ) with respect to ( x ):

[ f'(x) = 8x^7 ] [ f''(x) = 56x^{7-1} ] [ f''(x) = 56x^6 ]

Now, we evaluate the second derivative at the critical point ( x = 0 ):

[ f''(0) = 56(0)^6 ] [ f''(0) = 0 ]

Since ( f''(0) = 0 ), the second derivative test is inconclusive.

To determine if ( x = 0 ) corresponds to a local maximum, minimum, or a point of inflection, we can analyze the behavior of the function around this point. Upon inspection, since ( f(x) = x^8 ) is an even-powered function, it is symmetric about the y-axis, and at ( x = 0 ), there is neither a local maximum nor a local minimum; rather, it is a point of inflection.

Therefore, for the function ( f(x) = x^8 ), there are no local maximum or minimum points. The critical point ( x = 0 ) corresponds to a point of inflection.

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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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