If the following function is continuous, then what is the value of b?
g(f)={2t^2+2t-24 if f≠3
{b if f=3
0
3
7
14
None of these
If the following function is continuous, then what is the value of a?
h(f)={2t+b if f<0
{2cos(f)-3 if 0≤f≤(pi/2)
{asin(f)+5b if f>(pi/2)
0
1
2
4
whats the answer? ive been waiting 7 years for the answer
Whats the freaking answer
I got 0
The answer is 2
The second one is 2
To determine the value of b in the function g(f), we need to ensure that the function is continuous. In this case, for g(f) to be continuous, the value of b when f = 3 should match the value of g(f) when f ≠ 3.
Given the function g(f):
g(f) = {2t^2 + 2t - 24 if f ≠ 3
{b if f = 3
To make the function continuous, we set the value of g(f) to be the same when f ≠ 3 and f = 3:
2t^2 + 2t - 24 = b
We can solve this equation to find the value of b. Let's find the roots of the quadratic equation:
2t^2 + 2t - 24 = 0
We can factorize this equation as follows:
2(t + 6)(t - 2) = 0
Setting each factor equal to zero, we get:
t + 6 = 0 or t - 2 = 0
Solving these equations for t:
t = -6 or t = 2
Now substituting these values of t back into the equation to find b:
2(-6)^2 + 2(-6) - 24 = b or 2(2)^2 + 2(2) - 24 = b
Simplifying:
72 - 12 - 24 = b or 8 + 4 - 24 = b
36 = b or -12 = b
Therefore, the possible values for b are 36 or -12. Neither of these values is provided among the answer choices (0, 3, 7, 14, or None of these), so we can conclude that the function g(f) is not continuous for any value of b in the given answer choices.
Next, let's determine the value of a in the function h(f). Similar to the previous question, we need to satisfy the conditions for continuity:
Given the function h(f):
h(f) = {2t + b if f < 0
{2cos(f) - 3 if 0 ≤ f ≤ (pi/2)
{asin(f) + 5b if f > (pi/2)
To make the function continuous, we set the value of h(f) to be the same at the transition points:
2t + b = 2cos(f) - 3 and 2cos(f) - 3 = asin(f) + 5b
Now, let's consider the first equation:
2t + b = 2cos(f) - 3
Setting t = 0 and f = 0 in this equation, we have:
0 + b = 2cos(0) - 3
b = 2 - 3
b = -1
Therefore, the value of b is -1.
Now, the second equation becomes:
2cos(f) - 3 = asin(f) + 5b
Setting f = pi/2 in this equation, we have:
2cos(pi/2) - 3 = asin(pi/2) + 5(-1)
0 - 3 = 1 + (-5)
-3 = -4
This equation is not satisfied. As a result, we cannot find a value of a that would make the function h(f) continuous for the given answer choices (0, 1, 2, or 4). Therefore, None of these is the correct answer.
In summary:
- For the function g(f), b can be either 36 or -12, but these values are not provided among the answer choices.
- For the function h(f), there is no value of a that would make it continuous for the given answer choices.