# On what intervals the following equation is concave up, concave down and where it's inflection point is (x,y) #f(x)=x^8(ln(x))#?

- if
#0 < x < e^(-15/56)# then#f# is concave down; - if
#x > e^(-15/56)# then#f# is concave up; #x=e^(-15/56)# is a (falling) inflection point

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To determine where the function ( f(x) = x^8 \ln(x) ) is concave up, concave down, and where its inflection point is, we need to analyze its second derivative ( f''(x) ):

[ f(x) = x^8 \ln(x) ]

First, find the first and second derivatives:

[ f'(x) = 8x^7 \ln(x) + x^7 ]

[ f''(x) = 8 \cdot 7x^6 \ln(x) + 8x^6 \cdot \frac{1}{x} + 7x^6 ]

[ f''(x) = 8x^6(7 \ln(x) + 1) + 7x^6 ]

To find where the function is concave up and concave down, we look for where ( f''(x) > 0 ) and ( f''(x) < 0 ), respectively.

For ( f''(x) > 0 ):

[ 7 \ln(x) + 1 > 0 ]

[ \ln(x) > -\frac{1}{7} ]

[ x > e^{-1/7} ]

For ( f''(x) < 0 ):

[ 7 \ln(x) + 1 < 0 ]

[ \ln(x) < -\frac{1}{7} ]

[ x < e^{-1/7} ]

Thus, ( f(x) ) is concave up on ( (e^{-1/7}, \infty) ) and concave down on ( (0, e^{-1/7}) ).

To find the inflection point, we need to locate where the concavity changes, i.e., where ( f''(x) = 0 ) or is undefined.

[ 7 \ln(x) + 1 = 0 ]

[ \ln(x) = -\frac{1}{7} ]

[ x = e^{-1/7} ]

Therefore, the inflection point of ( f(x) ) is at ( (e^{-1/7}, f(e^{-1/7})) ).

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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.

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