Write the Lagrange formula and find xi for: (a) f(x) =

Write the Lagrange formula and find $\xi$ for: (a) $f(x) = 1-x^{2}$ on $[0,1]$; (b) $f(x) = \arctan x$ on $[0,1]$; (c) $f(x) = \ln x$ on $[1,e]$.

Step 1: The formula

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there is a $\xi\in(a,b)$ with

\[ f(b)-f(a) = f'(\xi)\bigl(b-a\bigr). \]

All three functions below are elementary and smooth on the intervals given, so the hypotheses need no further comment.

Step 2: Part (a): $f = 1-x^{2}$ on $[0,1]$

\[ f(1)-f(0) = 0-1 = -1, \qquad f'(x) = -2x, \]

\[ -1 = -2\xi\cdot1 \;\Longrightarrow\; \xi = \frac12 \in \bigl(0,1\bigr). \]

Step 3: Part (b): $f = \arctan x$ on $[0,1]$

\[ f(1)-f(0) = \frac{\pi}{4}, \qquad f'(x) = \frac{1}{1+x^{2}}, \]

\[ \frac{\pi}{4} = \frac{1}{1+\xi^{2}} \;\Longrightarrow\; 1+\xi^{2} = \frac{4}{\pi} \;\Longrightarrow\; \xi = \sqrt{\frac{4-\pi}{\pi}}, \]

taking the positive root since $\xi\in(0,1)$. Numerically $\xi \approx \sqrt{0.27324} \approx 0.52272$.

Step 4: Part (c): $f = \ln x$ on $[1,e]$

\[ f(e)-f(1) = 1, \qquad f'(x) = \frac1x, \]

\[ 1 = \frac{1}{\xi}\bigl(e-1\bigr) \;\Longrightarrow\; \xi = e-1 \approx 1.71828. \]

Step 5: Verification

\[\text{(a) } f'\left(\tfrac12\right) = -1 = \frac{f(1)-f(0)}{1} \;\;\checkmark\]

\[\text{(b) } \frac{1}{1+0.27324} = \frac{1}{1.27324} = 0.78540 = \frac{\pi}{4} \;\;\checkmark\]

\[\text{(c) } \frac{1}{e-1} = 0.58198 \text{ and } \frac{\ln e-\ln 1}{e-1} = \frac{1}{1.71828} = 0.58198 \;\;\checkmark\]