Explain why the 2nd derivative test is unable to classify all the critical numbers of h(s) = s 4 +4s 3 −1

    both the first and second derivative have zeros (horizontal axis intercepts when you graph the derivatives out)  at  . so there is a critical number at  , but since the second derivative is also 0 there, we cannot use the second derivative test.at a point where the second derivative is equal to zero, it is neither concave down nor concave up.graphically, the second derivative can do different things after it intercepts the horizontal axis. it could bounce off the horizontal axis  like a parabola and thus not change the function's concavity (making it a possible max/min). or it could go through the horizontal axis, making the second derivative change concavity (inflection point).consider  . the first and second derivatives are both at , but there is a local (and absolute) minimum at . its second derivative is  which bounces off the x-axis/horizontal axis (it is a parabola).consider  . the first and second derivatives are both at , but there is  an inflection point  (concavity changes); it does not have an extremum there. its second derivative is  , which  goes through the x-axis.there's not enough information to tell. have to use the first derivative test for classifying the critical number of at