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 has exactly one zero in :
- is strictly decreasing in , meaning it is one-to-one and crosses each value only once.
- and , so by the Intermediate Value Theorem (IVT), there must be a value where .
- Since the function is decreasing, there can be only one such zero.
Related Questions:
- How do we formally define differentiability at a point?
- What is the significance of the second derivative in determining concavity?
- How does the Intermediate Value Theorem guarantee at least one zero?
- Why does strict monotonicity ensure a unique zero?
- How can we modify the function to introduce additional points of non-differentiability?
Tip: When analyzing function properties, always check continuity before differentiability!
Ask a new question for Free
By Image
Drop file here or Click Here to upload
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
Related Recommendation
Analyzing a Graph Using Derivatives and Critical Points in Business Calculus
Graphing a Function with Derivative and Concavity Conditions
Sketching the Graph of a Continuous Function with Derivatives
Sketching f(x) from f'(x) for [0,8] with f(0) = 0
Solving a Function Derivative Problem with Symmetry and Inequality Constraints