12.19 Theorems of Derivative

Lagrange Mean Value Theorem

Let \(f:[a,b]\rightarrow \R\) is continuous and differentiable on \((a,b)\), then \(\exists c\in(a,b):f^{\prime}(c)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\)

Main idea: Apply rolle theorem

Proof

Take \(h(x)=f(x)-[\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)]\)

Apply Rolle's Theorem: \(h\) is continuous on \([a,b]\) and differentiable on \((a,b)\), Since \(h(a)=f(a)-f(a)=0,h(b)=f(b)-f(b)\), then we have \(h(a)=h(b)\).

Then by theorem, \(\exists c\in (a,b):h'(c)=0\), then we need \(h'(x)\)

\(h^{\prime}(x)=f^{\prime}\left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow h^{\prime}\left(c\right)=f^{\prime}\left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}=0\Rightarrow f^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\)

Proposition

If \(f:(a,b)\rightarrow \R\) and \(f\) is differentiable and \(f'(x)=0,\forall x\in(a,b)\), then \(\exists c\in \R:f(x)=c,\forall x\in (a,b)\)

Proof

Since we know if \(f(x)=c,\forall x\), then \(f'(x)=0,\forall x\)

Contradiction: Suppose \(f'(x)=0,\forall x\) and \(\exists x_1,x_2\in (a,b):f(x_1)\neq f(x_2)\)

Now, think \(f:[x_1,x_2]\rightarrow \R\) continuous and differentiable on \((x_1,x_2)\).

Apply mean value theorem, \(\exists z\in(x_1,x_2):f^{\prime}(z)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\neq0\), contradiction!

Thus \(f\) is a constant function on \((a,b)\)

Example

Tell all \(f:\R\to \R\) is differentiable such that \(f'(x)=x,\forall x\in \R\). Answer: \(f(x)=\frac{x^2}{2}+c\) for some \(c\in \R\)

Proof

If \(f'(x)=x,\forall x\), let \(g(x)=f(x)-\frac{x^2}{2}\), then \(g'(x)=0\), then \(g(x)=c\)

Then \(f(x)=\frac{x^2}{2}+c\)

Example

Tell all differentiable \(f:\R\to \R\) such that \(f'=f\). Answer: \(f(x)=ce^x\)

Proof

If \(f'(x)=f(x)\), let \(g(x)=\frac{f(x)}{e^x}\). Then \(g^{\prime}(x)=\frac{f^{\prime}\left(x\right)e^{x}-f\left(x\right)\left(e^{x}\right)^{\prime}}{e^{2x}}=\frac{f\left(x\right)e^{x}-f\left(x\right)e^{x}}{e^{2x}}=0\)

Then \(g(x)=c\), then \(f(x)=ce^x\)

Corollary

If differentiable \(f,g:(a,b)\to \R\) and \(f'=g'\), then \(\exists c\in \R\) such that \(f(x)=g(x)+c,\forall x\in (a,b)\)

Proof

If \(f^{\prime}(x)=g^{\prime}(x)\), let \(h(x)=f(x)-g\left(x\right)\). Then \(h^{\prime}(x)=f^{\prime}\left(x\right)-g^{\prime}\left(x\right)=0\)

Then \(h(x)=c\), then \(f\left(x\right)=g\left(x\right)+c\)

Corollary

If differentiable \(f:(a,b)\to \R\), then

If \(f'(x)>0,\forall x\in (a,b)\), then \(f\) is strictly monotonic increasing
If \(f^{\prime}(x)<0,\forall x\in(a,b)\), then \(f\) is strictly monotonic decreasing

Proof

Suppose \(f'(x)>0,\forall x\) but \(f\) is not monotonic increasing

Then \(\exists x_1<x_2:f(x_1)\geq f(x_2)\), then \(\exists z\in(x_1,x_2):f^{\prime}(z)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\leq0\)

Contradiction

Example

Plot \(f(x)=x^3-2x^2+x+1\)

Method to plot

\(\text{Dom}(f)\)
\(f(0),f(1),...\) some values roots. \(\lim_{x\to\pm\infty},\lim_{x\to a^{\pm}}\left(a\notin\text{dom}f(x)\right)\)(Asymptotes)
\(x-\)interception and \(y\)-interception(和x,y轴交点)
Symmetry? Even function and odd function
\(f'\), singular point, \(f'>0,f'<0\)
inflection(\(f''=0\)) \(f''>0\) \(f''<0\)
Connect with smooth curve

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