How do you solve #(x^2-9)*(x^2-16)<=0#?

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Answer 1 · 2015-06-22T22:45:45

#f(x) = (x^2-9)(x^2-16)# is a continuous function with zeros at #x=+-4# and #x=+-3# of multiplicity #1#.

Hence solution: #x in [-4, -3] uu [3, 4]#

Let #f(x) = (x^2-9)(x^2-16)#

#= (x-3)(x+3)(x-4)(x+4)#

#f(x)# is a continuous function with zeros at #x=+-4# and #x=+-3#.

Each of the roots of #f(x) = 0# is of multiplicity #1#, hence #f(x)# changes sign at those points.

For large positive and negative values of #x#, #f(x)# is large and positive.

Hence #f(x) <= 0# when #x in [-4, -3] uu [3, 4]#

Here's a graph of #f(x)/20#...

graph{(x^2-9)(x^2-16)/20 [-9.68, 10.32, -1.52, 8.48]}