Mathematics

Mathematics

Problem

Suppose ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b). Assume that there exists a constant k>0k>0 such that f′(x)≥kf'(x)\ge k for all xx in (a,b)(a,b). a. Show that for every xx and yy in [a,b][a,b] with x<yx<y, the inequality f(y)−f(x)≥k(y−x)f(y)-f(x)\ge k(y-x) holds. b. Deduce that ff is strictly increasing on [a,b][a,b] and, consequently, that the equation f(x)=cf(x)=c can have at most one solution in that interval for any real constant cc.
Intuitive explanation

The key insight is that a derivative bounded below by a positive constant tells you how fast the function must grow. Since f′(x)≥k>0f'(x) \geq k > 0 everywhere on the interval, the function is not just increasing—it's increasing at least as steeply as a line with slope kk. This means you can use the Mean Value Theorem or a direct integral argument to establish a lower bound on how much ff must change from aa to bb, which often leads to conclusions about injectivity, invertibility, or minimum growth rates.

Intuitive explanation

The key insight is that a derivative bounded below by a positive constant tells you how fast the function must grow. Since f′(x)≥k>0f'(x) \geq k > 0 everywhere on the interval, the function is not just increasing—it's increasing at least as steeply as a line with slope kk. This means you can use the Mean Value Theorem or a direct integral argument to establish a lower bound on how much ff must change from aa to bb, which often leads to conclusions about injectivity, invertibility, or minimum growth rates.

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