Let f(x)=(2x^2+5x-7)/(x-1)
show that F(x) has a removable discontinuity at x=1 and determine what value for F(1) would make f(x) continuous at x=1
I'm not sure how to factor to solve for f(1)..
factoring the numerator ... 2x^2+5x-7 = (2x + 7) (x - 1)
F(x) = 2x + 7 ... F(1) = ?
the discontinuity is zero divided by zero
... (x - 1) / (x - 1) for f(1)
f(x) and F(x) are very similar, with an important difference
f(x) = 2x+7
everywhere except at x=1.
so, if we define f(1) = 9 then f(x) will be continuous everywhere.
To determine if a function has a removable discontinuity at a specific point, we need to check if the function is undefined at that point, but can be made continuous by assigning a specific value to it.
In this case, we need to analyze the function f(x) = (2x^2 + 5x - 7)/(x - 1) at x = 1.
Let's start by factoring the numerator of f(x):
f(x) = (2x^2 + 5x - 7)/(x - 1)
The numerator can be factored as:
f(x) = [(2x - 1)(x + 7)]/(x - 1)
Now we can see that the function is undefined at x = 1 because it would result in a division by zero.
To determine the value of F(1) that would make f(x) continuous at x = 1, we need to check if there is a common factor between the numerator and the denominator that can cancel out the (x - 1) term.
Looking at the factored numerator, we see that there is a common factor of (x - 1):
f(x) = [(2x - 1)(x + 7)]/(x - 1)
Now we can cancel out the (x - 1) term:
f(x) = (2x - 1)(x + 7)
The function f(x) is now defined for all x ≠ 1 and we have eliminated the removable discontinuity at x = 1.
To find the value of F(1) that would make f(x) continuous at x = 1, we substitute x = 1 into the simplified function:
F(1) = (2(1) - 1)(1 + 7)
= (2 - 1)(8)
= 1(8)
= 8
Therefore, the value of F(1) that would make f(x) continuous at x = 1 is 8.