Suppose f(x)=x^p where p1 and [a,b]=[0,1]. According to the Mean Value Theorem there is at least one number c such that

Question

Suppose f(x)=x^p where p>1 and [a,b]=[0,1]. According to the Mean Value Theorem there is at least one number c such that
f(b)−f(a)=f′(c)(b−a).

In this particular case the number c is unique but it depends on p.

c=____

Your answer will be in terms of p.

Steve · Answer

f'(x) = p x^(p-1)
f(0)=0 and f(1)=1

So, you want c such that

p*c^(p-1) = 1

Now just solve for c. For example, if p=2,

f'(x) = 2x
2c = 1
c = 1/2

For p=3, 3c^2 = 1, c = 1/?3
so, the tangent line at x = 1/?3 is

y = (1/?3)^3 = 1(x - 1/?3)
or,
y = x - 1/?3 + 1/3^(3/2)

See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E3,+y+%3D+x-1%2F%E2%88%9A3+%2B+1%2F3%5E(3%2F2),+y%3Dx