How do you find the x values at which #f(x)=3x-cosx# is not continuous, which of the discontinuities are removable?

Question

Charles Arocha · Accepted Answer

#f(x)=3x-cosx# is continuous #AA x in RR#, and so it does not have any discontinuities.

#3x# is continuous for all real numbers #x# (we write as #AA x in RR# which means for all #x# in the set of real numbers). #cosx# is also continuous #AA x in RR# Therefore any linear combination of the above is also continuous #AA x in RR#. Hence #f(x)=3x-cosx# is continuous #AA x in RR#, and so it does not have any discontinuities. You can see this visually by looking at the graph of #y)=3x-cosx#, which is essentially that of #y=3x# with a slight oscillation caused by the addition of #-cosx# graph{3x-cosx [-30, 30, -30, 30]}

Isabella Ackerman · Answer

undefined To find the x values at which f(x)=3x-cosx is not continuous, we need to identify the points where the function has discontinuities.

The function f(x)=3x-cosx is a combination of a polynomial function (3x) and a trigonometric function (cosx).

The polynomial function 3x is continuous everywhere, so it does not contribute to any discontinuities.

The trigonometric function cosx is also continuous everywhere. However, it has discontinuities at the points where the denominator of the function becomes zero. In this case, the denominator is not present, so there are no discontinuities caused by the trigonometric function.

Therefore, the function f(x)=3x-cosx is continuous for all x values. There are no points of discontinuity, and thus, there are no removable discontinuities.