consider the function ( f(x)=8x + 5x^{-1} ).\nnote that this function has no inflection points, but ( f(x) )…

consider the function ( f(x)=8x + 5x^{-1} ).\nnote that this function has no inflection points, but ( f(x) ) is undefined at ( x = b ) where\n( b=)\nfor each of the following intervals, tell whether ( f(x) ) is concave up (type in cu) or concave down (type in cd).\n( (-infty,b):)\n( (b,infty):)

consider the function ( f(x)=8x + 5x^{-1} ).\nnote that this function has no inflection points, but ( f(x) ) is undefined at ( x = b ) where\n( b=)\nfor each of the following intervals, tell whether ( f(x) ) is concave up (type in cu) or concave down (type in cd).\n( (-infty,b):)\n( (b,infty):)

Answer

Explanation:

Step1: Find the first - derivative

Using the power rule ( (x^n)^\prime=nx^{n - 1} ), for ( f(x)=8x + 5x^{-1} ), we have ( f^\prime(x)=8-5x^{-2}=8-\frac{5}{x^{2}} ).

Step2: Find the second - derivative

Differentiate ( f^\prime(x) ) with respect to ( x ). Using the power rule again, ( f^{\prime\prime}(x)=10x^{-3}=\frac{10}{x^{3}} ).

Step3: Find where ( f^{\prime\prime}(x) ) is undefined

Set the denominator of ( f^{\prime\prime}(x) ) equal to zero. For ( y = \frac{10}{x^{3}} ), when ( x = 0 ), ( f^{\prime\prime}(x) ) is undefined. So ( B = 0 ).

Step4: Test intervals for concavity

  • For the interval ( (-\infty,0) ), let ( x=-1 ). Then ( f^{\prime\prime}(-1)=\frac{10}{(-1)^{3}}=-10<0 ). So ( f(x) ) is concave down (CD) on ( (-\infty,0) ).
  • For the interval ( (0,\infty) ), let ( x = 1 ). Then ( f^{\prime\prime}(1)=\frac{10}{1^{3}}=10>0 ). So ( f(x) ) is concave up (CU) on ( (0,\infty) ).

Answer:

( B = 0 ), ( (-\infty,B):CD ), ( (B,\infty):CU )