Question

The right-sided and left-sided derivatives of a function at a point $a$ are given by $f_{+}^{'} (a) = \lim_{h \to 0^{+}} \frac{f (a + h) - f (a)}{h}$ and $f_{-}^{'} (a) = \lim_{h \to 0^{-}} \frac{f (a + h) - f (a)}{h}$ , respectively, provided these limits exist. The derivative $f^{$\prime$}$\left$(a$\right$)$ exists if and only if $f_{+}^{$\prime$}$\left$(a$\right$)=f_{-}^{$\prime$}$\left$(a$\right$)$ .
Compute $f_{+}^{$\prime$}$\left$(a$\right$)$ and $f_{-}^{$\prime$}$\left$(a$\right$)$ at the given point $a$ .
$f (x) = {\begin{matrix} 4 - x^{2} if x \leq 1 \\ 2 x + 1 if x > 1 \end{matrix}$ ; $a equals 1$

Accepted Answer

Identify the piecewise function given: f(x) = 4 - $$x^2$$ for x ≤ 1 and f(x) = 2x + 1 for x \u003e 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 	o $$0^{-}$$} \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 	o $$0^{-}$$} \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 	o $$0^{+}$$} \frac{f(1+h) - f(1)}{h}. Since x \u003e 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 	o $$0^{+}$$} \frac{(2(1+h) + 1) - 3}{h}. Simplify the numerator and evaluate the limit to find f_{+}^{	ext{prime}}(1).

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Right-sided and Left-sided Derivatives

Existence of the Derivative

Piecewise Functions