Final Exam Prep

Function Defined: Consider the specific trigonometric function $k(t) = 432.971 \cos(91(t + 5)) + 921.584$ .
Amplitude ( $a$ ): * The amplitude is the absolute value of the coefficient of the trigonometric part of the function. * In this case, the exact amplitude is $a = 432.971$ .
Midline ( $y = c$ ): * The midline represents the horizontal line that the function oscillates around, determined by the vertical shift constant. * The equation for the midline is $y = 921.584$ .
Determining the Range: * The range defines all possible output values of the function, calculated from the minimum value to the maximum value. * Minimum value: Midline minus Amplitude ( $921.584 - 432.971 = 488.613$ ). * Maximum value: Midline plus Amplitude ( $921.584 + 432.971 = 1354.555$ ). * The range is represented in interval notation as $[488.613, 1354.555]$ .
Period ( $P$ ): * The period is the length of one complete cycle of the function. * It is calculated using the formula $P = \frac{2\pi}{b}$ , where $b$ is the coefficient of the input variable inside the trigonometric function. * Given $b = 91$ , the exact period is $P = \frac{2\pi}{91}$ .

Function Form: The function is defined as $f(t) = a e^{kt}$ .
Given Constraints: * At $t = 0$ , the output is $17$ ( $f(0) = 17$ ). * At $t = 3$ , the output is $16$ ( $f(3) = 16$ ).
Solving for Initial Value ( $a$ ): * Substitute the first point into the general form: $17 = a e^{k(0)}$ . * Since $e^0 = 1$ , the equation simplifies to $17 = a \times 1$ . * Therefore, the exact value of $a$ is $17$ .
Solving for Growth/Decay Constant ( $k$ ): * Using the value of $a$ and the second point $(3, 16)$ , set up the equation: $16 = 17 e^{k(3)}$ . * Isolate the exponential term: $\frac{16}{17} = e^{3k}$ . * Apply the natural logarithm to both sides to solve for the exponent: $\ln\left(\frac{16}{17}\right) = \ln(e^{3k})$ . * Using the inverse property: $\ln\left(\frac{16}{17}\right) = 3k$ . * Divide by $3$ to find the exact value: $k = \frac{1}{3} \ln\left(\frac{16}{17}\right)$ .
Justification: The value of $a$ represents the y-intercept of the function, while $k$ represents the continuous growth rate. Because f(3) < f(0), we expect $k$ to be negative, which is confirmed here since \ln(16/17) < 0.

Equation Provided: $3 \cdot 4^{(\frac{1}{2})t+2} = 5$
Isolation of the Exponential Base: * The first step is to isolate the term containing the variable by dividing both sides by the coefficient $3$ . * The equation becomes: $4^{(\frac{1}{2})t+2} = \frac{5}{3}$ .
Applying Logarithms: * Take the natural logarithm ( $\ln$ ) of both sides: $\ln\left(4^{(\frac{1}{2})t+2}\right) = \ln\left(\frac{5}{3}\right)$ .
Using Logarithmic Power Rules: * Bring the exponent down as a multiplier: $\left(\left(\frac{1}{2}\right)t + 2\right) \ln(4) = \ln\left(\frac{5}{3}\right)$ .
Isolating the Variable Term: * Divide by $\ln(4)$ : $\left(\frac{1}{2}\right)t + 2 = \frac{\ln(5/3)}{\ln(4)}$ . * Subtract $2$ from both sides: $\left(\frac{1}{2}\right)t = \frac{\ln(5/3)}{\ln(4)} - 2$ .
Final Solution: * Multiply the entire expression by $2$ to solve for $t$ : t = 2 \left(\frac{\ln(5/3)}{\ln(4)} - 2 ight). * Alternatively, the solution can be simplified to: $t = \frac{2 \ln(5/3)}{\ln(4)} - 4$ .

Initial Conditions: * Right Triangle: A triangle containing a $90^{\circ}$ ( $\pi/2$ radian) angle. * Given Angle: $\theta = \frac{\pi}{5}$ radians. * Given Side: Adjacent side length ( $L_{adj}$ ) = $17$ .
Solving for the Non-Right Angle (Complementary Angle): * In a right triangle, the two acute angles must sum to $\frac{\pi}{2}$ radians. * The second angle (let's call it $\phi$ ) is $\frac{\pi}{2} - \frac{\pi}{5}$ . * Converting to a common denominator: $\frac{5\pi}{10} - \frac{2\pi}{10} = \frac{3\pi}{10}$ . * The exact measure of the other angle is $\frac{3\pi}{10}$ radians.
Solving for the Opposite Side ( $L_{opp}$ ): * The relationship is defined by the tangent ratio: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$ . * $\tan\left(\frac{\pi}{5}\right) = \frac{L_{opp}}{17}$ . * The length of the opposite side is exactly $17 \tan\left(\frac{\pi}{5}\right)$ .
Solving for the Hypotenuse ( $L_{hyp}$ ): * The relationship is defined by the cosine ratio: $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$ . * $\cos\left(\frac{\pi}{5}\right) = \frac{17}{L_{hyp}}$ . * The exact length of the hypotenuse is $\frac{17}{\cos(\pi/5)}$ , which can also be written as $17 \sec\left(\frac{\pi}{5}\right)$ .

Identifying Zeros and Multiplicities: * Zeros: These are the x-intercepts of the polynomial $p(x)$ . * Determining Multiplicity: * If the graph crosses the x-axis linearly, the multiplicity is odd (typically $1$ ). * If the graph touches the x-axis and turns back (reflects), the multiplicity is even (typically $2$ ). * If the graph flattens out while crossing, the multiplicity is an odd number greater than $1$ (e.g., $3$ ).
Turning Points: * A turning point is a local maximum or minimum where the function changes its direction. * To identify these, one must estimate the x-coordinate for every peak and valley visible on the graph.
Determining the Least Possible Degree: * The degree of a polynomial relates to its number of roots and turning points. * Method 1: Sum the multiplicities of all identified zeros. * Method 2: Identify the number of turning points ( $n$ ). The least possible degree is typically $n + 1$ . * Justification Statement: One sentence explaining the choice—for example, "The degree must be at least $d$ because the function has $n$ turning points and its end behavior suggests an [even/odd] degree."
Assumptions: The analysis assumes that the provided graph displays all global behavior, including all zeros and relative extrema.