Determine all values of x for which each function is discontinuous.
F(x)= 3x+4 / 4x^2 - 2x -2
I have a feeling you meant to say
f(x) = (3x+4)/(4x^2 - 2x - 2)
which is
(3x+4)/(2(2x^2 - x - 1))
= (3x+4)/((2(2x + 1)(x - 1))
This is discontinuous at x = -1/2 and x = 1
Ah, discontinuities, the party animals of mathematics! Let's see what we can find for F(x).
To find where F(x) is discontinuous, we'll need to take a closer look at the denominator. F(x) will be discontinuous whenever the denominator (4x^2 - 2x - 2) equals zero, because dividing by zero is not allowed – it's like trying to divide a pizza by nothing, you end up with a sad situation!
So, let's solve 4x^2 - 2x - 2 = 0:
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a,
where a = 4, b = -2, and c = -2.
Plugging these values into the formula, we get:
x = (-(-2) ± √((-2)^2 - 4 * 4 * (-2))) / (2 * 4)
= (2 ± √(4 + 32)) / 8
= (2 ± √36) / 8
= (2 ± 6) / 8
So, the solutions for x are:
x = (2 + 6) / 8 = 8 / 8 = 1
x = (2 - 6) / 8 = -4 / 8 = -1/2
Therefore, F(x) is discontinuous at x = 1 and x = -1/2. Just make sure not to throw any wild pizza parties at these points!
To determine the values of x for which the function F(x) is discontinuous, we need to identify any values of x where there are vertical asymptotes, holes, or jumps in the graph of the function.
Vertical asymptotes occur when the denominator of a rational function (a function in the form f(x) = p(x)/q(x), where p(x) and q(x) are polynomials) equals zero.
In this case, the denominator of F(x) is 4x^2 - 2x - 2. To find the values of x for which the denominator equals zero, we need to solve the equation 4x^2 - 2x - 2 = 0.
Factoring this equation, we get:
(2x + 2)(2x - 1) = 0
Setting each factor equal to zero and solving for x, we find:
2x + 2 = 0 => 2x = -2 => x = -1
2x - 1 = 0 => 2x = 1 => x = 1/2
Therefore, the values of x for which the denominator equals zero are x = -1 and x = 1/2.
These values represent vertical asymptotes or holes in the graph of F(x), so F(x) is discontinuous at x = -1 and x = 1/2.
To determine the values of x for which a function is discontinuous, we need to investigate two cases:
1. The denominator is equal to zero.
2. The function is undefined at a particular x.
Let's start with the first case:
1. Denominator equal to zero:
In the given function, we have a quadratic denominator: 4x^2 - 2x - 2. We can solve this quadratic equation by factoring or using the quadratic formula.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4, b = -2, and c = -2.
x = (-(-2) ± √((-2)^2 - 4(4)(-2))) / (2(4))
x = (2 ± √(4 + 32)) / 8
x = (2 ± √36) / 8
x = (2 ± 6) / 8
This gives us two possible values for x:
x1 = (2 + 6) / 8 = 8 / 8 = 1
x2 = (2 - 6) / 8 = -4 / 8 = -1/2
So, we have two potential discontinuity points where the denominator is zero: x = 1 and x = -1/2.
Now, let's move on to the second case:
2. Undefined points:
The function can also be undefined at values of x that make the numerator equal to zero. In this case, the numerator is 3x + 4.
Setting the numerator equal to zero:
3x + 4 = 0
3x = -4
x = -4/3
So, we have one more potential discontinuity point where the numerator is zero: x = -4/3.
Combining the results from both cases, the values of x for which the function F(x) = (3x + 4) / (4x^2 - 2x - 2) is discontinuous are:
x1 = 1
x2 = -1/2
x3 = -4/3