The function f(x) = x² is given. At some point x
The function $f(x) = x^{2}$ is given. At some point $x$ the increment of the independent variable is $\Delta x = 0.2$, and the corresponding differential is $df = 0.8$. Find the value of $x$.
Explanation of Method
Reading the definition of the differential backwards.
Step 1: The definition
\[ df = f'(x)\,\Delta x. \]
Everything in this relation is known except $x$, so it becomes an equation for $x$.
Step 2: Substitute the function
\[ f'(x) = 2x \;\Longrightarrow\; df = 2x\,\Delta x. \]
Step 3: Insert the data and solve
\[ 2x\cdot0.2 = 0.8 \;\Longrightarrow\; 0.4\,x = 0.8 \;\Longrightarrow\; x = 2. \]
Step 4: Verification
\[f'(2) = 4, \qquad df = 4\times0.2 = 0.8 \;\;\checkmark\]
\[\text{Exact increment: } \Delta f = f(2.2)-f(2) = 4.84-4 = 0.84\]
\[\Delta f-df = 0.04 = \bigl(\Delta x\bigr)^{2}, \text{ exactly as for any quadratic} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,x = 2\,}\]
Remark
Remark. The equation has a unique solution here only because $f' = 2x$ is one-to-one. For a function whose derivative repeats values — $f(x) = x^{3}-3x$, say — the same data would admit several points, and for one with a bounded derivative the data might admit none at all. The problem is really asking where a given slope occurs.