The right-sided and left-sided derivatives of a function at a point a are given by and , respectively, provided these limits exist. The derivative f′(a) exists if and only if f+′(a)=f−′(a). Compute f+′(a) and f−′(a) at the given point a. ;
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Identify the piecewise function given: f(x) = 4 - x^2 for x ≤ 1 and f(x) = 2x + 1 for x > 1. We need to compute the right-sided and left-sided derivatives at a = 1.
Compute the left-sided derivative f_{-}^{ ext{prime}}(1) using the limit definition: f_{-}^{ ext{prime}}(1) = \lim_{h \to 0^{-}} \frac{f(1+h) - f(1)}{h}. Since x ≤ 1, use f(x) = 4 - x^2. Substitute f(1) = 4 - 1^2 = 3 and f(1+h) = 4 - (1+h)^2.
Simplify the expression for the left-sided derivative: f_{-}^{ ext{prime}}(1) = \lim_{h \to 0^{-}} \frac{(4 - (1+h)^2) - 3}{h}. Expand (1+h)^2 to get 1 + 2h + h^2, and simplify the numerator.
Compute the right-sided derivative f_{+}^{ ext{prime}}(1) using the limit definition: f_{+}^{ ext{prime}}(1) = \lim_{h \to 0^{+}} \frac{f(1+h) - f(1)}{h}. Since x > 1, use f(x) = 2x + 1. Substitute f(1) = 3 and f(1+h) = 2(1+h) + 1.
Simplify the expression for the right-sided derivative: f_{+}^{ ext{prime}}(1) = \lim_{h \to 0^{+}} \frac{(2(1+h) + 1) - 3}{h}. Simplify the numerator and evaluate the limit to find f_{+}^{ ext{prime}}(1).
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