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(i) Verify the Mean Value Theorem for 
y=x^(3)-3x+1 on the interval 
[-2,2] as follows.
(a) Verify that the hypotheses of the Mean value theorem holds.
(b) Verify that the conclusion of the Mean value theorem holds.
(ii) Illustrate (i-(b)) by a graph involving the extreme chord.

(i) Verify the Mean Value Theorem for y=x33x+1 y=x^{3}-3 x+1 on the interval [2,2] [-2,2] as follows.\newline(a) Verify that the hypotheses of the Mean value theorem holds.\newline(b) Verify that the conclusion of the Mean value theorem holds.\newline(ii) Illustrate (i-(b)) by a graph involving the extreme chord.

Full solution

Q. (i) Verify the Mean Value Theorem for y=x33x+1 y=x^{3}-3 x+1 on the interval [2,2] [-2,2] as follows.\newline(a) Verify that the hypotheses of the Mean value theorem holds.\newline(b) Verify that the conclusion of the Mean value theorem holds.\newline(ii) Illustrate (i-(b)) by a graph involving the extreme chord.
  1. Check Continuity: Check if the function is continuous on the closed interval [2,2][-2, 2].\newliney=x33x+1y = x^3 - 3x + 1 is a polynomial, which is continuous everywhere.
  2. Check Differentiability: Check if the function is differentiable on the open interval (2,2)(-2, 2).\newliney=x33x+1y = x^3 - 3x + 1 is a polynomial, which is differentiable everywhere.
  3. Calculate Average Rate: Calculate the average rate of change of the function on the interval [2,2][-2, 2].\newlinef(2)=233(2)+1=86+1=3f(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3\newlinef(2)=(2)33(2)+1=8+6+1=1f(-2) = (-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1\newlineAverage rate of change = (f(2)f(2))/(2(2))=(3(1))/(4)=4/4=1(f(2) - f(-2)) / (2 - (-2)) = (3 - (-1)) / (4) = 4 / 4 = 1
  4. Find Derivative: Find the derivative of the function.\newlinef(x)=ddx(x33x+1)=3x23f'(x) = \frac{d}{dx} (x^3 - 3x + 1) = 3x^2 - 3
  5. Set Derivative Equal: Set the derivative equal to the average rate of change to find the value cc in (2,2)(-2, 2).3x23=13x^2 - 3 = 13x2=43x^2 = 4x2=43x^2 = \frac{4}{3}x=±43x = \pm\sqrt{\frac{4}{3}}x=±43x = \pm\frac{\sqrt{4}}{\sqrt{3}}x=±23x = \pm\frac{2}{\sqrt{3}}Since we are looking for a value in the interval (2,2)(-2, 2), we take x=23x = \frac{2}{\sqrt{3}}.
  6. Verify Value of c: Verify that this value of cc is within the interval (2,2)(-2, 2).23\frac{2}{\sqrt{3}} is approximately 1.151.15, which is between 2-2 and 22.
  7. Draw Graph: Draw the graph to illustrate the conclusion of the Mean Value Theorem. The graph should show the function y=x33x+1y = x^3 - 3x + 1, the secant line connecting the points (2,f(2))(-2, f(-2)) and (2,f(2))(2, f(2)), and the tangent line at x=23x = \frac{2}{\sqrt{3}} that is parallel to the secant line.

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