Answered: Find The Values Of A And B That Make F…

How to find a and b for a piecewise function to be continuous everywhere. When considering a function to find x values where the function is not continuous, two very common things to look for would be taking the square root of a negative number or dividing by zero.Problem 46 Hard Difficulty. Find the values of $ a $ and $ b $ that make $ f $ continuous everywhere. Don't play me for everywhere, and a factor of ten gets ten plus Tony being a minus sixteen being he wants to twelve rearranging gets us for being close to two.I'll let you finish. Does this give you enough clues to figure out what mistakes you may have made? Finding the measure of an angle in a figure?Making a and b continuous everywhere. The question asks "find the values of "a" and "b" that make fcontinuous everywhere (answer in fractions)f(x)= ax2 - bx +4if 2 is less than or equal to x and xis <33x-a+bif x isgreater than or equal to 3I'm not sure where to start?What type of answer am...Find all values a and b that make this function continuous everywhere. Help finding the values for $a$ and $b$ that make the function continuous and differentiable everywhere. 0. Is it correct to say that the function $f(x)$ is everywhere continuous even if left side limit doesn't match with right...

SOLVED:Find the values of a and b that make f con

Since f is continuous everywhere, it must be continuous at x= 2 and 3. Apply the rule of continuity at x=2,3 i.e Left hand limit=Right Hand Limit and get the values of a and b. Hope this helps. get answers with explanations. find similar questions.Find class notes for your course.Then find the value of f(0) so that the function f is continuous at x=0. View solution. Customize assignments and download PDF's. Make now. Learn with content. Watch learning videos, swipe through stories, and browse through concepts.The values of a and b to make the function f(x) continuous is a = 1/2 and b = 1/2. Find a,b so that f is continuos everywhere. First note that we need only work at x=2 and x=3 as f is continuous on `(-oo,2),(2,3),(3,oo)` .(Polynomials and rationals are continuous everwhere on their domains).

Find the values of a and b that make f continuous everywhere?

So the question is asking to calculate or find the value of a and b so that it would be fit or be continuous in the said equation in your problem, base on my calculation and further formulation, the value of A is 7/12 and B = 13/2. I hope you are satisfied with my answer and feel free to ask for more.Sal finds the limit of a piecewise function at the point between two different cases of the function. Video transcript. so we've got this function f of X that is piecewise continuous is defined over let's first think about the limit the limit of f of X as we approach 2 from the left from values lower than 2 well...Hello, I need to find b and c parameters for this function, but I can't handle the e number. Thank you in advance for your help. You want the 2nd limit to exist, not be undefined. So think about what you want x^2 - 6x + b to equal when x=4. Then make that happen by defining b appropriately.Given F(x) = ax+3, x>5 8 , x=5 x^2 + bx + a, x<5 Find a and b so that f(x) is continuous everywhere. Posted 2 months ago. » Performance Management. » Decision Making. » Incremental Analysis: Make or Buy, Special Order. » Pricing. » Budget Planning.Define continuity on an interval. State the theorem for limits of composite functions. We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.

Continuous way that the graph of f(x) has no gaps in the periods.

f(x) is a piecewise function because for each and every period, there are differing types of behaviors.

Lets look at the first habits.

(x2 - 4) / (x - 2)

x can't equal 2 due to the denominator being zero. That is why x=2 isn't incorporated in that domain. However, if we enlarge the numerator and cancel, it simplifies to (x + 2). So you'll say that f(x) is (x + 2) in the period (-∞, 2).

We wish to evaluation (x + 2) when x=1.99999. We will use the f(x) worth for the 2d piece.

(1.9999 + 2) = 3.9999

This manner once we go to the 2d piece, which begins at x=2, the f(x) price could be 4. We do this because the first piece must be attached to the 2d piece to ensure that it to be continuous. Think of it like this: as x gets closer to 2 from the left, f(x) gets nearer to 4. This is what we call a prohibit.

Now we cross directly to the 2nd and third pieces. We use the domain names of these pieces to find a and b and the value of (x + 2) at x=1.9999 to create an equation.

Second piece:

a(2)2 - b(2) + 3 = 4

4a - 2b + 3 = 4

4a - 2b = 1

Next, we will give the parabola a vertex. If the x coordinate of the vertex lies between 2 and 3, then the endpoint of the parabola might be f(x) = 4. This is as a result of it is symmetrical.

Now to make the 2nd piece and 3rd piece share the identical level, we equate them. They will percentage the same level at x=3.

a(3)2 - b(3) + 3 = 4

9a - 3b = 1

2(3) - a + b = 4

6 - a + b = 4

-a + b = -2

Now, we use those equations to solve for a and b.

9a - 3b = 1 eq1

-a - b = -2 eq2

Multiply eq2 by 9. Keep eq1.

9a - 3b = 1 eq1

-9a - 9b = -18 eq2

Add each equation to do away with the a terms.

-12b = -17

b = 17 / 12

The exchange this value of b into either of those equations to solve for a. Once you have got a and b, plug them into the purposes and take a look at them out on a graphing calculator by means of plotting the 3 items to see if the graph is continuous.

I'm hoping this is helping. Even I discovered this to be a super problem.