Values of C?

Question

Consider the function f(x) = 4 x^3 - 4 x on the interval [ -4 , 4 ].

 

Find the average or mean slope of the function on this interval. 

 

By the Mean Value Theorem, we know there exists at least one c in the open interval ( -4 , 4 ) such that f'( c) is equal to this mean slope. 

 

For this problem, there are two values of c that work. The smaller one is     ???       and the larger one is ????

Answer 1

First find the values of the function at the ends of the interval.  That will give you two points, and the mean slope is the slope of the line that connects them.

 

So when x = -4, 4x³ - 4x = 4(-4)³ - 4(-4) = -256 + 16 = -240

 

When x = 4, 4x³ - 4x = 4(4)³ - 4(4) = 256 - 16 = 240

 

So the end points of the curve in that interval are (-4, -240) and (4, 240). 

 

The slope of the line between those points is (240 - -240) / (4 - -4) = 480 / 8 = 60

 

Now at what values of x is the slope equal to 60?

 

First you have to find the derivative of the function to find an expression for its slope.

 

The derivative is 3*4x² - 4 = 12x² - 4

 

So we have to solve the equation 12x² - 4 = 60

 

12x² - 4 = 60

12x² = 64

x² = 64/12 = 16/3

 

So x = ±√(16/3) = ±4/√(1/3) = ±4 /(√3) /3 = ±(4/3) √3

 

So the smaller value is -(4/3) √3 and the larger value is (4/3) √3