How do you determine whether the function #f(x)=x^8(ln(x))# is concave up or concave down and its intervals?

#D_f=R^+#

#f'(x)=8x^7lnx+x^8 1/x=8x^7lnx+x^7=x^7(8lnx+1)#

#f''(x)=7x^6(8lnx+1)+x^7 8/x=7x^6(8lnx+1)+8x^6#

#f''(x)=x^6(56lnx+15)#

#f''(x)=0 <=> x^6(56lnx+15)=0 <=>#

#<=> (x^6=0 vv 56lnx+15=0)#

#x^6=0 <=> x=0 !in D_f#

#56lnx+15=0 <=> 56lnx=-15 <=> lnx=-15/56 <=>#

#<=> x=e^(-15/56) <=> x=1/root(56)(e^15) ~~ 0.765>0 in D_f#

#AAx in (0,1/root(56)(e^15)): f''(x)<0# function is concave down

#AAx in (1/root(56)(e^15), oo): f''(x)>0# function is concave up

Note: the sign of #f''(x)# depends only on #56lnx+15# since #x^6# is always greater then zero for #x>0#.

To determine the concavity of the function \( f(x) = x^8 \ln(x) \) and its intervals, we need to find the second derivative of the function and analyze its sign. 1. Find the first derivative \( f'(x) \) using the product rule. \[ f'(x) = 8x^7 \ln(x) + x^7 \cdot \frac{1}{x} \] \[ f'(x) = 8x^7 \ln(x) + x^6 \] 2. Find the second derivative \( f''(x) \) by differentiating \( f'(x) \). \[ f''(x) = 8 \cdot 7x^6 \ln(x) + 8x^6 \cdot \frac{1}{x} \] \[ f''(x) = 56x^6 \ln(x) + 8x^5 \] 3. To determine the concavity, set \( f''(x) \) equal to zero and find the critical points. \[ 56x^6 \ln(x) + 8x^5 = 0 \] At \( x = 0 \), the function is not defined, so we only consider positive values of \( x \). 4. Solve \( 56x^6 \ln(x) + 8x^5 = 0 \) for positive \( x \). \[ x^5(56x\ln(x) + 8) = 0 \] \( x^5 = 0 \) has no positive solutions, so we focus on \( 56x\ln(x) + 8 = 0 \). \[ 56x\ln(x) + 8 = 0 \] \[ 56x\ln(x) = -8 \] \[ \ln(x) = -\frac{8}{56x} \] \[ x = e^{-\frac{1}{7}} \] 5. Analyze the sign of \( f''(x) \) in intervals created by the critical point. - For \( x < e^{-\frac{1}{7}} \), \( f''(x) > 0 \) (Concave up) - For \( x > e^{-\frac{1}{7}} \), \( f''(x) < 0 \) (Concave down) Therefore, the function \( f(x) = x^8 \ln(x) \) is concave up for \( x < e^{-\frac{1}{7}} \) and concave down for \( x > e^{-\frac{1}{7}} \).