A: If the function f is continuous on [?2,1] and f(?2) = 1, f(0) = ?2 and f(1) =

Question

A: If the function f is continuous on [?2,1] and f(?2) = 1, f(0) = ?2 and f(1) = 0, is it possible that f is one-to-one? If the answer is yes, give an example of a one-to-one function satisfying these conditions. If the answer is no, provide a justification that any function satisfying these conditions cannot be one-to-one.

Explanation / Answer

Yes, the function is one to one. Let the function be ax^2 + bx +c = y Sub (-2,1) in the function => 4a - 2b + c = 1 Sub (0,-2) in the function => c = -2 Sub (1,0) in the function => a+b+c = 0 Solving the above equations, we get a=1.5 b = 0.5 c =-2 Therfore, the equation is y = 1.5x^2 + 0.5x - 2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.