Discussion. Given an initial value problemy′ + p(t)y = q(t), y(t0) = y0 ,the interval of validity is the largest interval that contain t0 and does not containany discontinuity for p(t) or q(t). The existence and uniqueness theorem guaranteesthere is a unique solution on the interval of validity.Problem: Without solving, determine the interval of validity for the initial valueproblem(t2 − 9)y′ + 2y = ln |20 − 4t| , y(4) = −3.

5. (a) Find the values of X₁ and x₂ in the figure. (b) Find the largest positive number & such that: |-1|<0.1 whenever 0<|x-1|<8. Hint: Choose & to be the minimum value of |x-1| for i=1,2. 1=1+0.1- 1- 9 -1-0.1- 10 + x₁ 1 || x2

State the Intermediate Value Theorem. Use it to show that the equationf (x) =x5 +10x +1 =0has a solution in the interval (−1, 1). Make sure you verify that the hypotheses of the theoremhold in this case and explain how the theorem is used. Then find an interval of length lessthan 0.1 that contains a solution of f (x) =0.

3. Match the rule or formula's name with its characteristics by writing the lowercase letter on the line. M(b − a)³ 24n² A. Midpoint Rule B. Trapezoidal Rule C. Simpson's Rule D. Error bound for Midpoint Rule E. Error bound for Trapezoidal Rule F. Error bound for Simpson's Rule a. Error < Ax b. - (ƒ(xo) + 4ƒ (x₁) + 2ƒ (x₂)+...+ 4f(xn-1) + f(n)) 3 c. Error d. Σ f(m;)Δα i=1 e. Ax 2 M (b − a)5 180n4 f. Error < where M = max(f"(x)) -(f(xo)+2f(ı)+2f(2)+ ...+2f(xn-1) + f(n)) M(b-a)³ 12n² " where M = max(|ƒ(4)(x)|) where M = max(f"(x))