Show that the function f(x)=1-|x|, [-1,1] does not satisfy the hypotheses of the mean value theorem on the given interval.
Also how do I graph the function together with the line through the points
A(a,f(a)) and B(b,f(b)).
Also how do I find values of c in (a,b) that satisfy f'(c)=f(b)-f(a) / b-a
Thank you so much in advance.
The function is not differentiable at 0. The assumption of the MVT is that is is continous, and differentiable on the interval.
Points a will have to be in the same side of origin as point b ie, either -1 to 0, or 0 to 1. Given that, then c will be between a and b.
To show that the function f(x) = 1 - |x| does not satisfy the hypotheses of the mean value theorem on the interval [-1, 1], we need to verify two conditions:
1. Continuity: The function should be continuous on the interval [a, b].
2. Differentiability: The function should be differentiable on the open interval (a, b).
Let's first consider continuity. The given function f(x) = 1 - |x| is indeed continuous except at the point x = 0. The function has a corner or "kink" at x = 0, which means it fails to meet the condition of continuity at that point. Therefore, f(x) is not continuous on the interval [-1, 1].
Secondly, to examine the differentiability, we need to find the derivative of f(x). Let's consider x < 0 and x > 0 separately:
For x < 0: f(x) = 1 - (-x) = 1 + x.
The derivative of f(x) = 1 + x is f'(x) = 1.
For x > 0: f(x) = 1 - x.
The derivative of f(x) = 1 - x is f'(x) = -1.
Notice that the derivative of f(x) is not defined at x = 0 since we have different values for x < 0 and x > 0. This means the function f(x) is not differentiable at x = 0.
Hence, since the function fails to satisfy both the continuity and differentiability conditions on the interval [-1, 1], it does not satisfy the hypotheses of the mean value theorem on that interval.
Moving on to graphing the function f(x) = 1 - |x| and the line passing through the points A(a, f(a)) and B(b, f(b)), you can approach it by following these steps:
1. Plot the points A(a, f(a)) and B(b, f(b)) on the coordinate plane.
2. Draw a straight line passing through A and B. This line represents the secant line between the two points. You can determine the slope of the line using the formula (f(b) - f(a)) / (b - a).
3. Plot the points along the function f(x) = 1 - |x| on the same coordinate plane.
4. Finally, draw the graph of the function f(x) = 1 - |x|. You can use the characteristics of the function to guide you, such as the fact that it is symmetric about the y-axis and has a corner at x = 0.
To find the values of c in the interval (a, b) that satisfy f'(c) = (f(b) - f(a)) / (b - a), you need to evaluate the derivative of f(x) and set it equal to the slope of the secant line.
For the given function f(x) = 1 - |x|, we know that the derivative is not defined at x = 0. So, we need to consider values of c in the interval (a, b) excluding c = 0.
Once you find the derivative of f(x), equate it to the slope of the secant line [(f(b) - f(a)) / (b - a)]. Solve the equation to find the values of c located between a and b.
Note: Keep in mind that since the function is not differentiable at x = 0, there may be multiple values of c that satisfy the equation.
I hope this explanation helps. Let me know if you have any further questions!