Precalculus Assignment_ Functions and Trigonometry

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May 26, 2024

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Precalculus Assignment: Functions and Trigonometry Instructions: Complete the following problems related to functions, trigonometry, and complex numbers. Show all work for full credit. Use proper mathematical notation and ensure your answers are clear and well-organized. Problem 1: Analyzing Functions 1. Question: Given the function f(x)=2x3−5x2+4x−1f(x) = 2x^3 - 5x^2 + 4x - 1f(x)=2x3−5x2+4x−1, find: a. The y-intercept. b. The zeros of the function (roots). c. The intervals where the function is increasing and decreasing. Answer: a. Y-intercept: To find the y-intercept, set x=0x = 0x=0: f(0)=2(0)3−5(0)2+4(0)−1=−1f(0) = 2(0)^3 - 5(0)^2 + 4(0) - 1 = -1f(0)=2(0)3−5(0)2+4(0)−1=−1 The y-intercept is (0,−1)(0, -1)(0,−1). b. Zeros of the function: To find the zeros, solve f(x)=0f(x) = 0f(x)=0: 2x3−5x2+4x−1=02x^3 - 5x^2 + 4x - 1 = 02x3−5x2+4x−1=0 Using synthetic division or numerical methods (since this is a cubic polynomial), the roots are approximately x≈0.5,1,2x \approx 0.5, 1, 2x≈0.5,1,2. c. Intervals of increase and decrease: First, find the derivative f′(x)f'(x)f′(x): f′(x)=6x2−10x+4f'(x) = 6x^2 - 10x + 4f′(x)=6x2−10x+4 Set the derivative equal to zero to find critical points: 6x2−10x+4=0   x=1±136x^2 - 10x + 4 = 0 \implies x = 1 \pm \frac{1}{\sqrt{3}}6x2−10x+4=0 x=1±31 Test intervals around the critical points: For x<1−13x < 1 - \frac{1}{\sqrt{3}}x<1−31, f′(x)>0f'(x) > 0f′(x)>0 (increasing). For 1−13<x<1+131 - \frac{1}{\sqrt{3}} < x < 1 + \frac{1}{\sqrt{3}}1−31<x<1+31, f′(x)<0f'(x) < 0f′(x)<0 (decreasing). For x>1+13x > 1 + \frac{1}{\sqrt{3}}x>1+31, f′(x)>0f'(x) > 0f′(x)>0 (increasing). Problem 2: Trigonometric Identities
2. Question: Prove the identity: sin 2(x)+cos 2(x)=1\sin^2(x) + \cos^2(x) = 1sin2(x)+cos2(x)=1. Answer: Start with the Pythagorean identity from trigonometry: sin 2(x)+cos 2(x)=1\sin^2(x) + \cos^2(x) = 1sin2(x)+cos2(x)=1 This identity is derived from the Pythagorean Theorem applied to the unit circle, where the hypotenuse is 1. For any angle xxx, the coordinates (cos (x),sin (x))(\cos(x), \sin(x))(cos(x),sin(x)) form a right triangle with the hypotenuse 1, thus proving the identity. Problem 3: Complex Numbers 3. Question: Simplify the expression (3+4i)(2−i)(3 + 4i)(2 - i)(3+4i)(2−i). Answer: Use the distributive property to multiply the complex numbers: (3+4i)(2−i)=3 2+3 (−i)+4i 2+4i (−i)(3 + 4i)(2 - i) = 3 \cdot 2 + 3 \cdot (-i) + 4i \cdot 2 + 4i \cdot (-i)(3+4i)(2−i)=3 2+3 (−i)+4i 2+4i (−i) Simplify each term: =6−3i+8i−4i2= 6 - 3i + 8i - 4i^2=6−3i+8i−4i2 Since i2=−1i^2 = -1i2=−1: =6−3i+8i+4=10+5i= 6 - 3i + 8i + 4 = 10 + 5i=6−3i+8i+4=10+5i Therefore, the simplified expression is 10+5i10 + 5i10+5i. Problem 4: Exponential Functions 4. Question: Solve for xxx in the equation 52x+1=1255^{2x+1} = 12552x+1=125. Answer: First, express 125 as a power of 5: 125=53125 = 5^3125=53 Therefore, the equation becomes: 52x+1=535^{2x+1} = 5^352x+1=53 Since the bases are the same, set the exponents equal to each other: 2x+1=32x + 1 = 32x+1=3 Solve for xxx: 2x=2   x=12x = 2 \implies x = 12x=2 x=1 Thus, x=1x = 1x=1.
Extra Credit: 5. Question: Verify the angle sum identity for sine: sin (A+B)=sin Acos B+cos Asin B\sin(A + B) = \sin A \cos B + \cos A \sin Bsin(A+B)=sinAcosB+cosAsinB. Answer: To verify this identity, consider the unit circle and the definition of sine and cosine as projections on the x and y axes. Using the sum of angles formula derived from the unit circle or trigonometric addition formulas, we get: sin (A+B)=sin Acos B+cos Asin B\sin(A + B) = \sin A \cos B + \cos A \sin Bsin(A+B)=sinAcosB+cosAsinB This can be verified by considering specific values for AAA and BBB, or by using geometric proofs involving the unit circle and properties of right triangles.
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