First, find the derivative using the power rule:

Now, substitute 3 for x.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of

 on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of an exponential: 

Trigonometric derivative: 

Using a solver, there are multiple solutions which satisfy this equation:

However, to satisfy the mean value theorem, a solution must fall within the specified interval.  falls within .

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our interval and function definitions, along with the derivative value of  given, we'll in turn wish to solve for  to validate the mean value theorem:

(Note that the solution  would not work because it would eliminate the interval as , and it would lead to a zero denominator in the working equation.)

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, , , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  gives the solution:

Which is, as expected, greater than  to allow the creation of an interval.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, ,  , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  gives the solution:

Which is, as expected, greater than  to allow the creation of an interval.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, , , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  gives the solution:

Which is, as expected, greater than  to allow the creation of an interval.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, , , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  via numerical solver gives the solution:

Choosing the positive solution for the sake of the interval leaves

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, , , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  using a calculator gives the solution:

Which is, as expected, greater than  to allow the creation of an interval.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, , , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  using a calculator gives the solution:

Which is, as expected, greater than , albeit just barely, to allow the creation of an interval.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Using our function, interval, and derivative definitions, , , , we'll in turn wish to solve for  to validate the mean value theorem:

Solving for  using a calculator gives the solution:

There are two solutions to the equation; elect the value larger than  to allow creation of an interval: