Precalculus Assignment_ Functions and Trigonometry
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North Carolina School of Science and Mathematics *
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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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