g (x)= (x^2 + 4x - 5) / (x^2 + 7x + 10)= ( (x-1) (x+5))/ ( (x+2) (x+5)) Domain x!=-2 and x!=-5 as this would make the denominator 0 As long as x!=-5 we may cancel out the (x+5)'s We get g (x)= …

Feb 27, 2015 · The only "forbidden" values for x are the ones that make the denominator equal to 0. (2x+2)/ (x^2-6x-7)= (2 (x+1))/ ( (x+1) (x-7) The domain is now limited to x!=-1 and x!=7 …

If a graph is drawn for 1/x as shown below, there is a discontinuity at zero. As for very small negative numbers, 1/x approaches negative infinity, and for very small positive numbers, 1/x …

Explanation: In fact, there is a family of functions #f# which are never continuous in any point, and their square is continuous: they are called the Dirichlet functions: fix #n>0# an integer, then:

Apr 30, 2018 · An Intuitive "proof": The function is defined for all values #x# with the exception of a discontinuity at #x=0#. Hence, given the restricted domain #x in (0,oo)# the function is well …

Futhermore #x≠0# should be a removable discontinuity, and would show up as a hole on the graph, rather than a vertical asymptote.

Q&A and Videos that make learning easy. Math, History, Chemistry, Algebra, Calculus, Biology and more.

How do you sketch the graph by determining all relative max and min, inflection points, finding intervals of increasing, decreasing and any asymptotes given #f (x)=x-x^ (2/3) (5/2-x)#?

Q&A and Videos that make learning easy. Math, History, Chemistry, Algebra, Calculus, Biology and more.

Q&A and Videos that make learning easy. Math, History, Chemistry, Algebra, Calculus, Biology and more.