consider the function ( f(x)=4x + 9x^{-1} ). note that this function has no inflection points, but ( f(x) )…

consider the function ( f(x)=4x + 9x^{-1} ). note that this function has no inflection points, but ( f(x) ) is undefined at ( x = b ) where ( b=square ) for each of the following intervals, tell whether ( f(x) ) is concave up (type in cu) or concave down (type in cd). ( (-infty,b):square ) ( (b,infty):square )

consider the function ( f(x)=4x + 9x^{-1} ). note that this function has no inflection points, but ( f(x) ) is undefined at ( x = b ) where ( b=square ) for each of the following intervals, tell whether ( f(x) ) is concave up (type in cu) or concave down (type in cd). ( (-infty,b):square ) ( (b,infty):square )

Answer

Explanation:

Step1: Find the first - derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (y = f(x)=4x + 9x^{-1}), we have (f^\prime(x)=\frac{d}{dx}(4x)+\frac{d}{dx}(9x^{-1})). (f^\prime(x)=4-9x^{-2}=4-\frac{9}{x^{2}}).

Step2: Find the second - derivative

Differentiate (f^\prime(x)) with respect to (x). Using the power rule again, (f^{\prime\prime}(x)=\frac{d}{dx}(4)-\frac{d}{dx}(9x^{-2})). (f^{\prime\prime}(x)=0 + 18x^{-3}=\frac{18}{x^{3}}).

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

The function (y = f^{\prime\prime}(x)=\frac{18}{x^{3}}) is undefined when (x = 0), so (B = 0).

Step4: Test the concavity on the intervals

  • For the interval ((-\infty,0)): Let (x=-1), then (f^{\prime\prime}(-1)=\frac{18}{(-1)^{3}}=-18<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{18}{1^{3}}=18>0). So (f(x)) is concave up (CU) on ((0,\infty)).

Answer:

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