Given f(x) = -e^(0.1x(1-x)), compute f(1.05) approximately
Given $f(x) = -e^{0.1x(1-x)}$, compute $f(1.05)$ approximately.
Explanation of Method
A base point chosen so that the exponent vanishes.
Step 1: Choose the base point
Write $1.05 = 1+0.05$, so $x_{0} = 1$ and $\Delta x = 0.05$. The choice is forced by the exponent: the factor $1-x$ vanishes at $x = 1$, which makes both $f(1)$ and $f'(1)$ trivial to evaluate.
\[ f(1.05) \approx f(1)+f'(1)\cdot0.05. \tag{1} \]
Step 2: The value at $x_{0} = 1$
\[ g(x) = 0.1\,x\bigl(1-x\bigr), \qquad g(1) = 0.1\cdot1\cdot0 = 0, \qquad f(1) = -e^{0} = -1. \]
Step 3: The derivative at $x_{0} = 1$
By the chain rule $f'(x) = -e^{g(x)}g'(x)$, and $g(x) = 0.1x-0.1x^{2}$ gives
\[ g'(x) = 0.1\bigl(1-2x\bigr), \qquad g'(1) = -0.1, \]
\[ f'(1) = -e^{0}\cdot\bigl(-0.1\bigr) = +0.1. \]
Step 4: Apply the linearisation
\[ f(1.05) \approx -1+0.1\times0.05 = -1+0.005 = -0.995. \]
Step 5: Verification
\[g(1.05) = 0.1\cdot1.05\cdot\bigl(1-1.05\bigr) = 0.1\cdot1.05\cdot\bigl(-0.05\bigr) = -0.00525\]
\[f(1.05) = -e^{-0.00525} = -0.9947638 \text{ exactly}\]
\[\text{The estimate } -0.995 \text{ is low by } 0.000236 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,f(1.05) \approx -0.995\,}\]
Remark
Remark. The discrepancy comes from the quadratic term of $g$, not from the exponential: $g(1.05) = -0.00525$ while its linear part is $g'(1)\cdot0.05 = -0.005$, a difference of $0.00025$ — essentially the whole error, since $e^{u}\approx1+u$ contributes only $u^{2}/2 \approx 1.4\cdot10^{-5}$ more. Note also the shape of $f$: the exponent $0.1x(1-x)$ is a downward parabola peaking at $0.025$ when $x = \tfrac12$, so $f = -e^{g}$ attains its minimum $-e^{0.025} \approx -1.0253$ there and rises towards $0^{-}$ as $|x|$ grows.