Prove Monotony of Function: $f$ Strictly Decreasing

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SUMMARY

The discussion focuses on proving that a differentiable function $f$, defined on the interval $(-\infty, 0)$ to $(0, \infty)$, is strictly decreasing given that $f'(x) < 0$ for all real numbers except at $x = 0$, where $f'(0) = 0$. The proof employs a contradiction approach, starting with the definition of a decreasing function and utilizing the mean value theorem to establish that if $f$ were not strictly decreasing, it would contradict the established derivative conditions.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and the mean value theorem.
  • Knowledge of the definition of strictly decreasing functions.
  • Familiarity with proof techniques, particularly proof by contradiction.
  • Basic concepts of real analysis, including function behavior and continuity.
NEXT STEPS
  • Study the mean value theorem in detail, including its applications and implications.
  • Explore the properties of differentiable functions and their derivatives.
  • Investigate proof techniques in mathematics, focusing on proof by contradiction.
  • Review examples of strictly decreasing functions and their graphical representations.
USEFUL FOR

This discussion is beneficial for mathematics students, particularly those studying calculus and real analysis, as well as educators looking for clear examples of function behavior and proof strategies.

awsomeman
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Let $f$ be differentiable from $(-\inf,0)$ to $(0,\inf)$ and let $f'(x)<0$ for all real numbers except 0 and $f'(0)=0$. Prove that f is strictly decreasing.
 
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You might want to begin by stating the definition of a decreasing function. Then consider some examples.
 
I would use "proof by contradiction". Suppose f is NOT strictly decreasing. Then there exist a, b, b> a, such that f(b)\ge f(a). So $f(b)- f(a)\ge 0$. Since b> a, b- a> 0 so $\frac{f(b)- f(a)}{b- a}\ge 0$. Now use the "mean value" property.
 

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