1. f(x) 3- {sqr(x) where x is greater than or equal to 4
{x^2+3 where x is less than 4
Plug 4 into each equation, approaching the left and right
Equation #1 comes to 19
Equation #2 comes to 1
They are not equal, therefore f(x) is not continuous
2. f(x)=2x^2+5x-3 [0,4]
Plug 0 and 4 into the equation
f(0) comes out to equal -3
f(4) comes out to equal 49
Since one is negative and the other is positive, there is indeed a solution
f(x)= -3x^2+14x-8 [-3,0]
Plug -3 and 0 into the equation
f(-3) comes out to equal -77
f(0) comes out to equal -8
Since they are both negative, a solution cannot be proven.
3. Two types - Different quotient and slope of a tangent line
Linear derivative - f(x)=2x-3 as x approaches 1
2x-3- f(1) / x-1
2x-3 - 1 / x-1
2x-2 / x-1
(x-1)(x+1) / x-1
x+1 = 2
A derivative is the slope of a line at a certain point
The hardest part is the difference quotient and doing the algebraic part of distributing the xs and hs.
4. Instantaneous velocity is the slope of one certain point. Average velocity is the average slope of a point.
For number 4, even though your definition is right, I would suggest adding an example to demonstrate more than just the dictionary definition of instantaneous velocity and average velocity. It will help show that you know how to find it on a quiz.
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ReplyDeleteOn number 2, instead of just saying "Indeed there is a solution" I think you should write it out like Christian showed us "If f is continuous in the interval [0,4], and f(a) < 0 <f(b) then there exists c on the [0,4} where f(c) = 0. I just don't want my friend miss points on a test for not writing the answer out correctly. ;*
ReplyDeleteYeah Trint's right. Why can't a solution be proven on 2b? Because we can't say for sure if it crossed the x axis or not, so we don't know if it has a solution.
ReplyDeleteOn 3, you may want to say what the difference quotients formula is and say it is when h approaches 0.
ReplyDeleteI agree with Swanson. You should also provide an example or solve your current example with the difference quotient.
ReplyDeleteThey are right. You should watch your wording on IVT. You should say that "a solution cannot be proven based on the interval with IVT". You might be able to provide a solution on a different interval or with another theorem. Just keep that in mind.
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