Math Problem Statement

Solve this and give me direct answer

Solve the following: 1. Let f(x) = x^2 - x: (a) Find an interval on which f satisfies the hypotheses of Rolle's Theorem. (b) Find all values of c that satisfy Rolle's Theorem. 3. Let f(x) = x^2 - x: (a) Find a point b such that the slope of the secant line through (0, 0) and (b, f(b)) is 1. (b) Find all values of c that satisfy the Mean-Value Theorem on [0, b].

Solution

Solution for Exercise 5.3:

1. Let f(x)=x2xf(x) = x^2 - x:

(a) An interval on which ff satisfies the hypotheses of Rolle's Theorem:

To satisfy Rolle's Theorem:

  1. ff must be continuous on [a,b][a, b].
  2. ff must be differentiable on (a,b)(a, b).
  3. f(a)=f(b)f(a) = f(b).

Choose the interval [0, 1] since:

  • f(x)f(x) is continuous and differentiable everywhere (as it's a polynomial).
  • f(0)=f(1)=0f(0) = f(1) = 0, satisfying f(a)=f(b)f(a) = f(b).

Thus, the interval is:
[0, 1].


(b) Find all values of cc that satisfy Rolle's Theorem:

Rolle's Theorem states f(c)=0f'(c) = 0 for some c(0,1)c \in (0, 1).

  1. Compute f(x)=2x1f'(x) = 2x - 1.
  2. Solve f(c)=0f'(c) = 0: 2c1=0    c=12.2c - 1 = 0 \implies c = \frac{1}{2}.

Thus, the value of cc is:
c=12c = \frac{1}{2}.


3. Let f(x)=x2xf(x) = x^2 - x:

(a) Find a point bb such that the slope of the secant line through (0,0)(0, 0) and (b,f(b))(b, f(b)) is 1:

The slope of the secant line is: slope=f(b)f(0)b0=1.\text{slope} = \frac{f(b) - f(0)}{b - 0} = 1.

Since f(0)=0f(0) = 0, this simplifies to: b2bb=1.\frac{b^2 - b}{b} = 1.

Simplify: b1=1    b=2.b - 1 = 1 \implies b = 2.

Thus, the point is:
b=2b = 2.


(b) Find all values of cc that satisfy the conclusion of the Mean-Value Theorem (MVT) for ff on [0,b][0, b], where b=2b = 2:

The Mean-Value Theorem states f(c)=f(b)f(0)b0f'(c) = \frac{f(b) - f(0)}{b - 0} for some c(0,b)c \in (0, b).

  1. Compute the slope of the secant line: f(2)f(0)20=22202=22=1.\frac{f(2) - f(0)}{2 - 0} = \frac{2^2 - 2 - 0}{2} = \frac{2}{2} = 1.

  2. Solve f(c)=1f'(c) = 1: f(x)=2x1    2c1=1    c=1.f'(x) = 2x - 1 \implies 2c - 1 = 1 \implies c = 1.

Thus, the value of cc is:
c=1c = 1.


Final Answers:

  1. (a) Interval: [0, 1]
    (b) Value of cc: c=12c = \frac{1}{2}

  2. (a) Point bb: b=2b = 2
    (b) Value of cc: c=1c = 1


Would you like detailed steps or have any other questions?

Here are 5 related questions to consider:

  1. What happens if f(x)f(x) has discontinuities in the interval?
  2. How does Rolle's Theorem relate to the Mean-Value Theorem?
  3. Can a cubic function satisfy Rolle's Theorem?
  4. What if f(a)f(b)f(a) \neq f(b) for some interval—does MVT still apply?
  5. How can we geometrically interpret the value of cc in both theorems?

Tip: Always verify the continuity and differentiability conditions before applying Rolle's or Mean-Value Theorems.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Differentiation
Critical Points
Continuity

Formulas

Slope of a line = (f(b) - f(a)) / (b - a)
Derivative f'(x) = 2x - 1

Theorems

Rolle's Theorem
Mean-Value Theorem

Suitable Grade Level

Grades 11-12