# The Mean Value Theorem applies to the given function on the given interval. How do you find all possible values of #f(x)=x^(9/4)# on the interval [0,1]?

The value of

The Mean Value Theorem states that :

Here,

Therefore,

Taking the natural logs on both sides

graph{x^(9/4) [-1.698, 2.626, -0.227, 1.936]}

By signing up, you agree to our Terms of Service and Privacy Policy

To find all possible values of ( f(x) = x^{9/4} ) on the interval ([0, 1]) using the Mean Value Theorem, we first find the derivative of the function, then find the critical points within the interval, and finally evaluate the function at these critical points.

The derivative of ( f(x) = x^{9/4} ) is given by:

[ f'(x) = \frac{9}{4}x^{5/4} ]

Now, we need to find critical points within the interval ([0, 1]) by setting the derivative equal to zero and solving for ( x ):

[ \frac{9}{4}x^{5/4} = 0 ]

Since ( x ) cannot be zero in the interval ([0, 1]), the only critical point occurs when the derivative is undefined, which is at ( x = 0 ).

Now, we evaluate the function ( f(x) = x^{9/4} ) at the endpoints of the interval ([0, 1]) and at the critical point ( x = 0 ):

- At ( x = 0 ), ( f(0) = 0^{9/4} = 0 ).
- At ( x = 1 ), ( f(1) = 1^{9/4} = 1 ).

So, the possible values of ( f(x) = x^{9/4} ) on the interval ([0, 1]) are ( 0 ) and ( 1 ).

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find all local maximum and minimum points given #y=x^4-2x^2+3#?
- Please help #f(x)=6x^5-10x^3# a. find the #x# coordinates of all max and min points. b. State the intervals where f is increasing?
- How do you find the critical points(using partial derivative) for #f( x , y )=x^3 + y^2 - 6x^2 + y -1#?
- What are the absolute extrema of # f(x)= |sin(x) - cos(x)|# on the interval [-pi,pi]?
- What are the critical values, if any, of #f(x) = (x - 1) / (x + 3) -sqrt(x^2-3)#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7