AP Calculus Math Problem PLEASE HELP!!!

Test Preparation AP Test Preparation
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<p>Please do not use your calculator!
Let f be the function defined as follows:</p>

<p>(a) If a = 2 and b = 3, is f continuous at x = 1? Justify your answer.</p>

<p>(b) Find a relationship between a and b (hint: a relationship between a and b just means an equation in a and b) for which f is continuous at x = 1.</p>

<p>(c) Find a relationship between a and b so that f is continuous at x = 2</p>

<p>(d) Use your equations from parts (b) and (c) to find the values of a and b so that f is continuous at both x = 1 and also at x = 2?</p>

<p>Please submit answers within 24 hours. Thanks so much!!</p>

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<p>How does a and b define function f?</p>

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<p>No homework on CC, right?</p>

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<p>i agree with drosser. I don't see how you hve enough info to answer the question. I'm sure you just forgot to write the functions a and b right?</p>

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<p>
[quote]
Please submit answers within 24 hours. Thanks so much!!

[/quote]

Not a good way to ask for help if you ask me...
have you attempted the problem?</p>

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<p>I'm pretty sure that f would be a function of x with coefficients a and b.</p>

<p>That being said, if you know the definition of continuity, you should be able to answer these questions.</p>

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<p>Sorry guys...</p>

<p>f(x)= abs(x-1) + 2, for x<1
ax^2+bx, for 1 less than or equal to x < 2
5x-10 for x is greater than or equal to 2</p>

<p>Please help me out I have no idea how to even start this.</p>

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<p>okay then,
(a) At x=1, |x-1| + 2 = 2, and 2x^2+3x = 5, so no, f is not continuous because |x-1|+2 is not equal to 2x^2+3x at x =1. It should be continuous only if the pieces of the function pick up where the function last left off, i.e. |x-1|+2 = ax^2+3x at x=1</p>

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<p>(b) For the function to be continuous, as I said above, it must pick up where it left off, as in no sudden jumps. So at x = 1, |x-1|+2 = ax^2+bx. We substitute 1 for x, so we have |1-1| +2=a+b, and now we have 2=a+b, so the relationship we seek is 2-a=b, or 2-b=a
(c)Same as b, except we substitute in 2 for x and set ax^2+bx=5x-10
Now we have 4a+2b=10-10, so 4a+2b=0
Which gives us 2b=-4a, or 4a=-2b, and to simplify even further, -2a=b or a=-0.5b</p>

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<p>(d) Now for the last part, we find two values for a and b that satisfy the equations we found in (b) and (c). That is, for the system of equations
-2a=b and
2-b=a , the values of a and b we pick must be so that both equations are true.
We can do this by substitution, so since -2a=b
2-2a=a, and solving for a, we get 2/3=a. Plugging that back into one equation, we get -2(2/3)=b, so b=-4/3</p>

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<p>Whoops, it seems I messed up on the signs. We actually have 2-(-2a)=a, so
2+2a=a and so -2=a. plugging that again, we get 2-b=-2, and so b=4</p>

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<p>Finally, we check our answer to see that it satisfies the conditions, so at x=1, |x-1|+2 = -2x^2+4x. Plugging in x=1. we get 0+2=-2+4, and so 2=2, satisfying the first condition.</p>

<p>Now we check the second condition, which is
-2x^2+4x=5x-10 at x=2
Plugging in x=2, we have
-2(2)^2+4(2)=5(2)-10
giving us
0=0
which checks off the second condition.</p>

<p>You're better off if you did this step too. As you can see I messed up an earlier calculation and wouldn't have caught that had I not done this step.</p>

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<p>Thanks so much It'sGr82BeAGator! You're such a lifesaver!!!</p>

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Hey for (b) I was just wondering why you didn’t use the 5x +10

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Ahhh, nevermind i figured it out !