The function is , .

Mean value theorem :

Let  be a function that satisfies the following three hypotheses.

1.  is continuous on .

2.  is differentiable on .

Then there is a number  in  such that  .

The function is .

The function is continuous on the interval .

Differentiate  on each side with respect to .

The function is differentiable on the interval .

Then .

From the mean value theorem :

Substitute  in .

Find the tangent line at .

From above consider .

Then .

Therefore the tangent line at a point is .

The slope of the tangent is the derivative of the function, .

Point - slope form of a line equation : .

Substitute  in the point - slope form.

The tangent line equation is .

From above consider .

Then .

Therefore the tangent line at a point is .

The slope of the tangent is the derivative of the function, .

Point - slope form of a line equation with a point  : .

Substitute  in the point - slope form.

The tangent line equation is .

Find the secant line through the end points.

The end points are  and .

.

.

Therefore the end points are  and .

Slope of the secant line is .

Point - slope form of a line equation : .

Substitute  in the point - slope form.

The secant line equation is .

Graph :

Graph the function  :

(1) Draw the coordinate plane.

(2) Draw the tangent line equation is  and .

(3) Draw the secant line equation is .

The tangent line equation are  and .

The secant line equation is .

Graph of the function, tangent lines and secant line on same graph :

.