1. **Problem:** Find the tangent line to the function $f(x) = x^2 - 4x + 1$ at the point where $x=0$. Step 1: Calculate $f(0)$. $$f(0) = 0^2 - 4\cdot0 + 1 = 1$$ Step 2: Find the derivative $f'(x)$. $$f'(x) = 2x - 4$$ Step 3: Calculate the slope at $x=0$. $$f'(0) = 2\cdot0 - 4 = -4$$ Step 4: Write the equation of the tangent line using point-slope form: $$y - f(0) = f'(0)(x - 0)$$ $$y - 1 = -4x$$ $$y = -4x + 1$$ --- 2. **Problem:** Find the tangent line to $f(x) = x^4 + \sqrt{x^2 + 1}$ at $x=0$. Step 1: Calculate $f(0)$. $$f(0) = 0^4 + \sqrt{0^2 + 1} = 1$$ Step 2: Find the derivative $f'(x)$. $$f'(x) = 4x^3 + \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = 4x^3 + \frac{x}{\sqrt{x^2 + 1}}$$ Step 3: Calculate the slope at $x=0$. $$f'(0) = 4\cdot0^3 + \frac{0}{\sqrt{0 + 1}} = 0$$ Step 4: Equation of tangent line: $$y - 1 = 0(x - 0)$$ $$y = 1$$ --- 3. **Problem:** Find the tangent line to $f(x) = \frac{x^2 + 3x - 4}{2}$ at $x=0$. Step 1: Calculate $f(0)$. $$f(0) = \frac{0 + 0 - 4}{2} = -2$$ Step 2: Find the derivative $f'(x)$. $$f'(x) = \frac{2x + 3}{2}$$ Step 3: Calculate the slope at $x=0$. $$f'(0) = \frac{0 + 3}{2} = \frac{3}{2}$$ Step 4: Equation of tangent line: $$y - (-2) = \frac{3}{2}(x - 0)$$ $$y + 2 = \frac{3}{2}x$$ $$y = \frac{3}{2}x - 2$$ --- 4. **Problem:** Find the tangent line to $f(x) = \frac{2 - \sin^4 x}{\cos x + 3}$ at $x=0$. Step 1: Calculate $f(0)$. $$f(0) = \frac{2 - \sin^4 0}{\cos 0 + 3} = \frac{2 - 0}{1 + 3} = \frac{2}{4} = \frac{1}{2}$$ Step 2: Use quotient rule for derivative: $$f'(x) = \frac{(\cos x + 3)(-4\sin^3 x \cos x) - (2 - \sin^4 x)(-\sin x)}{(\cos x + 3)^2}$$ Step 3: Evaluate numerator and denominator at $x=0$: - $\sin 0 = 0$, $\cos 0 = 1$ - Numerator: $$ (1 + 3)(-4 \cdot 0 \cdot 1) - (2 - 0)(-0) = 4 \cdot 0 - 2 \cdot 0 = 0$$ - Denominator: $$ (1 + 3)^2 = 16$$ Step 4: Calculate slope at $x=0$: $$f'(0) = \frac{0}{16} = 0$$ Step 5: Equation of tangent line: $$y - \frac{1}{2} = 0(x - 0)$$ $$y = \frac{1}{2}$$