Graph the function below over the indicated interval hx x3x
     Graph the function below over the indicated interval.  h(x) = x(3^x); [-4.0]  Choose the correct graph below.  ![Graph the function below over the indicated interval. h(x) = x(3^x); [-4.0] Choose the correct graph below. SolutionOn the given interval: x<=0 and 3^x>0  Graph the function below over the indicated interval. h(x) = x(3^x); [-4.0] Choose the correct graph below. SolutionOn the given interval: x<=0 and 3^x>0](/WebImages/11/graph-the-function-below-over-the-indicated-interval-hx-x3x-1007029-1761519247-0.webp) 
  
  Solution
On the given interval: x<=0 and 3^x>0
So, h(x)<=0
At x=0,h(0)=0
As magnitude of x increases 3^x goes to 0 and since 3^x goes to 0 faster than x goes to -infinity the whole graph goes to 0
A,B,C seem to fit this description.
So we now look at second derivatives to find curvature of graph as x goes to -infinity
h\'=3^x+x3^x log(3)
h\'\'=2log(3) 3^x + x3^x log(3)^2=3^x log(3)(2+x log (3))
So we see there is inflection point at:x=-2/log(3) hence, A and C seem to fit the description.
the critical point ie where derivative is 0 is x=-1/log(3)
At this point the value of function is
h(-log(3))~-0.3
Hence, C is correct option
![Graph the function below over the indicated interval. h(x) = x(3^x); [-4.0] Choose the correct graph below. SolutionOn the given interval: x<=0 and 3^x>0  Graph the function below over the indicated interval. h(x) = x(3^x); [-4.0] Choose the correct graph below. SolutionOn the given interval: x<=0 and 3^x>0](/WebImages/11/graph-the-function-below-over-the-indicated-interval-hx-x3x-1007029-1761519247-0.webp)
