Remember that every polynomial is continuous everywhere, and the quotient of two continuous functions is continuous everywhere except where the denominator is zero. Also, the absolute value function is continuous and the composition of two continuous functions is also continuous.
Note that here we have the (continuous) absolute value function composed with the function `(5x^2+12x-9)/(8x+24)` , which is continuous everywhere except for `x=-3`, where it isn't even defined. So using what we said in the first paragraph, our only possible point of discontinuity for the function `f` is `x=-3`, and we need to compute the limit of `f(x)` as `x` approaches `-3`.
First, factor and cancel as follows:
`f(x)=|(5x^2+12x-9)/(8x+24)|=|((5x-3)(x+3))/(8(x+3))|=|(5x-3)|/8` , whenever `x!=-3`.
Now, when `x->-3` , it's clear that `f(x)->18/8=9/4`. But by definition, `f(-3)=9/4` . Since `lim_(x->-3)f(x)=f(-3)` , `f` is continuous at `x=-3`, and so it is continuous everywhere.
`f` is continuous for all values of `x`.