Given that f(x)= 2x^2+3x-20/ x+4 , find the value(s) of the "x" causing f(x) to be discontinuous and the limit of f(x) as X approaches these value(s)
thanks guys for your help!
2007-10-27 10:03:34 · 4 answers · asked by jasmine 2 in Science & Mathematics ➔ Mathematics
f(x)= (2x^2+3x-20)/ (x+4)
f(x) is discontinuous when x = -4 because the denominator is zero fot that value of x.
When x --> -4 f(x) --> 0/0 which is indeterminate
Use L'Hospital's rule getting:
lim x ---> -4 (4x+3)/1= -13
2007-10-27 10:13:38 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
Discontinuities can appear for several reasons, but the most common is a division by zero. If you look at your problem for a place you would find such a discontinuity, you should find that if x = 4 you are in trouble with a divide by zero.
Now you want to find that limit. This is possible by numerical substitution of a number like 3.9999999 into the formula to get 41.5, but if you need a derivation of the limit we can go there as well, just indicate it in your question.
2007-10-27 17:21:52 · answer #2 · answered by jestephen 2 · 0⤊ 0⤋
f(x) = (2x^2 + 3x - 20) / (x + 4)
f(x) is discontinuous at x = -4
as x approaches -4, f(x) becomes 0/0 form.
Applying L'Hospital's rule for limit
Lim f(x) as x -> -4 is 4(-4) + 3 = -13
2007-10-27 17:12:40 · answer #3 · answered by psbhowmick 6 · 0⤊ 0⤋
the graph is discountinuous when there is division by zero at x = -4.
2007-10-27 17:12:45 · answer #4 · answered by Terry S 3 · 0⤊ 0⤋