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.
To find the number c in terms of p, we can use the Mean Value Theorem. According to the theorem, if a function f(x) is continuous on the interval [a, b] and differentiable on the interval (a, b), then there exists at least one number c in the interval (a, b) such that:
f(b) - f(a) = f'(c)(b - a)
In this case, the function f(x) = x^p is continuous on the interval [0, 1] and differentiable on the interval (0, 1) for any p > 1. To find c, we need to find the derivative of f(x) and substitute it into the above equation.
First, let's find the derivative of f(x):
f'(x) = p*x^(p-1)
Now, substitute f'(x) = p*x^(p-1) into the Mean Value Theorem equation:
f(b) - f(a) = f'(c)(b - a)
(b^p) - (a^p) = p*c^(p-1)(b - a)
Since a = 0 and b = 1, we have:
(1^p) - (0^p) = p*c^(p-1)(1 - 0)
1^p - 0 = p*c^(p-1)
Simplifying further, we have:
1 = p*c^(p-1)
To solve for c, we need to isolate it. Divide both sides of the equation by p:
1/p = c^(p-1)
Now, take the p-1 power of both sides:
(1/p)^(1/(p-1)) = c
Therefore, the value of c in terms of p is:
c = (1/p)^(1/(p-1))
This gives you the unique value of c for the given function f(x) = x^p and the interval [0, 1].
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