Given f(x) = cases x², x < 1,, ax+b, x > 1, cases
Given $f(x) = \begin{cases} x^{2}, & x \lt 1, \\ ax+b, & x \gt 1, \end{cases}$ determine $a$ and $b$ so that $f$ is differentiable at $x = 1$.
Explanation of Method
Two conditions — matching values and matching slopes — fix two constants.
Step 1: The continuity condition
Differentiability at a point implies continuity there, so the one-sided limits must agree: $\lim_{x\to 1^{-}}x^{2} = 1, \qquad \lim_{x\to 1^{+}}(ax+b) = a+b$.
\[ \Longrightarrow\qquad a+b = 1. \tag{1} \]
This also fixes the value at the junction: $f(1) = 1$.
Step 2: The matching-slope condition
Compute the one-sided derivatives at $x = 1$ from the definition: \[f'_{-}(1) = \lim_{h\to 0^{-}}\frac{(1+h)^{2}-1}{h} = \lim_{h\to 0^{-}}\frac{2h+h^{2}}{h} = \lim_{h\to 0^{-}}(2+h) = 2,\]
\[ f'_{+}(1) = \lim_{h\to 0^{+}}\frac{a(1+h)+b-(a+b)}{h} = \lim_{h\to 0^{+}}\frac{ah}{h} = a. \]
Differentiability requires them to be equal: \[a = 2. \tag{2}\]
Step 3: Solve
From (1) and (2): $b = 1 - a = -1$.
Step 4: Verification
\[\text{Values: } \lim_{x\to 1^{-}}x^{2} = 1 = \lim_{x\to 1^{+}}(2x-1) \;\;\checkmark\]
\[\text{Slopes: } f'_{-}(1) = 2 = f'_{+}(1), \text{ so } f'(1) = 2 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,a = 2, \qquad b = -1 \qquad\Longrightarrow\qquad f(x) = \begin{cases} x^{2}, & x \leq 1, \\ 2x-1, & x > 1. \end{cases}\,}\]
Remark
Remark. The line $y = 2x-1$ is precisely the tangent to $y = x^{2}$ at $(1,1)$ — which is the geometric content of the answer: the only line that joins a parabola smoothly is its own tangent. Note that $f''$ still jumps at $x = 1$ (from $2$ to $0$), so the join is $C^{1}$ but not $C^{2}$; matching second derivatives too would need a quadratic on the right, not a line.