121616949-math.149 - 6.5 The Mean Value Theorem 135 So the...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
6.5 The Mean Value Theorem 135 So the height of g ( x ) is the same at both endpoints. This means, by Rolle’s Theorem, that at some c , g ( c ) = 0. But we know that g ( c ) = f ( c ) - m , so 0 = f ( c ) - m = f ( c ) - f ( b ) - f ( a ) b - a , which turns into f ( c ) = f ( b ) - f ( a ) b - a , exactly what we want. Returning to the original formulation of question (2), we see that if f ( t ) gives the position of your car at time t , then the Mean Value Theorem says that at some time c , f ( c ) = 70, that is, at some time you must have been traveling at exactly your average speed for the trip, and that indeed you exceeded the speed limit. Now let’s return to question (1). Suppose, for example, that two functions are known to have derivative equal to 5 everywhere, f ( x ) = g ( x ) = 5. It is easy to find such functions: 5 x , 5 x + 47, 5 x - 132, etc. Are there other, more complicated, examples? No—the only functions that work are the “obvious” ones, namely, 5 x plus some constant. How can we see that this is true?
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern