1. **Problem Statement:** Find the derivatives using the definition and rules of derivatives for the given functions and analyze differentiability and continuity. --- ### Q1: Using Definition of Derivative 2. **Show that** $f(x) = |x - 6|$ **is not differentiable at** $x=6$. - Recall the definition of derivative at $a$: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ - For one-sided derivatives: $$f'_-(a) = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}, \quad f'_+(a) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$$ 3. **Calculate one-sided derivatives at $x=6$:** - For $x < 6$, $f(x) = 6 - x$ so $$f'_-(6) = \lim_{h \to 0^-} \frac{|6 + h - 6| - 0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$$ - For $x > 6$, $f(x) = x - 6$ so $$f'_+(6) = \lim_{h \to 0^+} \frac{|6 + h - 6| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$$ 4. Since $f'_-(6) \neq f'_+(6)$, $f'(6)$ does not exist. Thus, $f$ is not differentiable at $x=6$. 5. **Formula for $f'(x)$:** $$f'(x) = \begin{cases} -1 & x < 6 \\ 1 & x > 6 \end{cases}$$ --- ### Q1 (a): Find $f'_-(4)$ and $f'_+(4)$ for $$f(x) = \begin{cases} 0 & x \leq 0 \\ 5 - x & 0 < x < 4 \\ \frac{1}{5 - x} & x \geq 4 \end{cases}$$ 6. Calculate left-hand derivative at $x=4$: - For $x < 4$, $f(x) = 5 - x$, so $$f'_-(4) = \lim_{h \to 0^-} \frac{(5 - (4 + h)) - 1}{h} = \lim_{h \to 0^-} \frac{1 - h - 1}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$$ 7. Calculate right-hand derivative at $x=4$: - For $x \geq 4$, $f(x) = \frac{1}{5 - x}$, so $$f'_+(4) = \lim_{h \to 0^+} \frac{\frac{1}{5 - (4 + h)} - 1}{h} = \lim_{h \to 0^+} \frac{\frac{1}{1 - h} - 1}{h} = \lim_{h \to 0^+} \frac{\frac{1 - (1 - h)}{1 - h}}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1 - h}}{h} = \lim_{h \to 0^+} \frac{1}{1 - h} = 1$$ --- ### Q1 (b): Sketch the graph of $f$: - Horizontal line $y=0$ for $x \leq 0$. - Decreasing line from $(0,5)$ to $(4,1)$ with slope $-1$. - Curve $y = \frac{1}{5 - x}$ for $x \geq 4$ with vertical asymptote at $x=5$. --- ### Q1 (c): Discontinuities - Check at $x=0$: $$\lim_{x \to 0^-} f(x) = 0, \quad \lim_{x \to 0^+} f(x) = 5 - 0 = 5$$ - Limits differ, so $f$ is discontinuous at $x=0$. - Check at $x=4$: $$\lim_{x \to 4^-} f(x) = 5 - 4 = 1, \quad \lim_{x \to 4^+} f(x) = \frac{1}{5 - 4} = 1$$ - Continuous at $x=4$. --- ### Q1 (d): Points of non-differentiability - At $x=0$ due to discontinuity. - At $x=4$ since $f'_-(4) = -1 \neq 1 = f'_+(4)$, not differentiable at $x=4$. --- ### Q2: Using Rule of Derivative 8. Find derivative of $$f(x) = (1 + 2x^2)(x - x^2)$$ - **Product Rule:** $$f'(x) = (1 + 2x^2)'(x - x^2) + (1 + 2x^2)(x - x^2)'$$ - Compute derivatives: $$(1 + 2x^2)' = 4x, \quad (x - x^2)' = 1 - 2x$$ - Substitute: $$f'(x) = 4x(x - x^2) + (1 + 2x^2)(1 - 2x)$$ - Expand: $$= 4x^2 - 4x^3 + 1 - 2x + 2x^2 - 4x^3 = 1 - 2x + 6x^2 - 8x^3$$ 9. **Multiply first:** $$f(x) = (1 + 2x^2)(x - x^2) = x - x^2 + 2x^3 - 2x^4$$ - Derivative: $$f'(x) = 1 - 2x + 6x^2 - 8x^3$$ - Both methods agree. --- ### Q2: Derivative of $$F(x) = \frac{x^4 - 5x^3 + \sqrt{x}}{x^2}$$ 10. **Quotient Rule:** - Let numerator $u = x^4 - 5x^3 + x^{1/2}$, denominator $v = x^2$. - Derivatives: $$u' = 4x^3 - 15x^2 + \frac{1}{2}x^{-1/2}, \quad v' = 2x$$ - Quotient rule: $$F'(x) = \frac{u'v - uv'}{v^2} = \frac{(4x^3 - 15x^2 + \frac{1}{2}x^{-1/2})x^2 - (x^4 - 5x^3 + x^{1/2})2x}{x^4}$$ - Simplify numerator: $$= (4x^5 - 15x^4 + \frac{1}{2}x^{3/2}) - (2x^5 - 10x^4 + 2x^{3/2}) = 2x^5 - 5x^4 - \frac{3}{2}x^{3/2}$$ - So $$F'(x) = \frac{2x^5 - 5x^4 - \frac{3}{2}x^{3/2}}{x^4} = 2x - 5 - \frac{3}{2}x^{-5/2}$$ 11. **Simplify first:** $$F(x) = x^2 - 5x + x^{-3/2}$$ - Derivative: $$F'(x) = 2x - 5 - \frac{3}{2}x^{-5/2}$$ - Both methods agree. --- **Final answers:** - $f'(6)$ does not exist for $f(x) = |x-6|$. - $f'_-(4) = -1$, $f'_+(4) = 1$ for piecewise $f$. - $f$ discontinuous at $x=0$, not differentiable at $x=0,4$. - Derivatives for $f(x)$ and $F(x)$ confirmed by two methods.