Math Problem Statement

Sketch a graph that is :

a) Differentiable on (-5,1) and (1,5) and not differentiable at 1.

b) f'(x)<0 on (-5,1)

c) f"(x)=0 on (-5,1), f"(x)>0 on (1,3), f"(x)<0 on (3,5)

d) f is continuous at x=1

e) f(-4)=3, f(-1)=-2

f) Prove that f has one and only one zero on the interval (-5,1).

Solution

Here is the sketch of the graph based on the given conditions. Let me know if you need any modifications or further explanations!


Proof that f(x)f(x) has exactly one zero in (5,1)(-5,1):

  • f(x)f(x) is strictly decreasing in (5,1)(-5,1), meaning it is one-to-one and crosses each value only once.
  • f(4)=3f(-4) = 3 and f(1)=2f(-1) = -2, so by the Intermediate Value Theorem (IVT), there must be a value c(5,1)c \in (-5,1) where f(c)=0f(c) = 0.
  • Since the function is decreasing, there can be only one such zero.

Related Questions:

  1. How do we formally define differentiability at a point?
  2. What is the significance of the second derivative in determining concavity?
  3. How does the Intermediate Value Theorem guarantee at least one zero?
  4. Why does strict monotonicity ensure a unique zero?
  5. How can we modify the function to introduce additional points of non-differentiability?

Tip: When analyzing function properties, always check continuity before differentiability!

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Math Problem Analysis

Mathematical Concepts

Differentiability
Continuity
First Derivative
Second Derivative
Intermediate Value Theorem
Monotonicity
Concavity

Formulas

f'(x) < 0
f''(x) = 0, f''(x) > 0, f''(x) < 0

Theorems

Intermediate Value Theorem
Monotonicity Theorem

Suitable Grade Level

Grades 11-12