Problem 37 - Algebra for College Students

Question

For Problems \(35-42\), (a) find the \(y\) intercepts, (b) find the \(x\) intercepts, and (c) find the intervals of \(x\) where \(f(x)>0\) and those where \(f(x)<0\). Do not sketch the graphs. $$ f(x)=(x+3)^{4}(x-1)^{3} $$

Step-by-Step Solution

Answer

y-intercept: (0, -81); x-intercepts: (-3, 0) and (1, 0); f(x)<0 on (-∞,-3)∪(-3,1); f(x)>0 on (1,∞).

Step 1: Find the y-intercept

To find the y-intercept of the function, we need to evaluate the function at \(x = 0\). This will give us the y-coordinate where the graph intersects the y-axis. Substitute into the function:\[f(0) = (0 + 3)^{4}(0 - 1)^{3} = 3^{4} imes (-1)^{3} = 81 imes (-1) = -81\]Thus, the y-intercept is \((0, -81)\).

Step 2: Find the x-intercepts

To find the x-intercepts, we set the function equal to zero and solve for \(x\). The function is zero when either factor \((x+3)^4\) or \((x-1)^3\) equals zero:\[(x+3)^4 = 0 \quad \Rightarrow \quad x = -3\]\[(x-1)^3 = 0 \quad \Rightarrow \quad x = 1\]Thus, the x-intercepts are \((-3, 0)\) and \((1, 0)\).

Step 3: Determine intervals where f(x) > 0 and f(x) < 0

To determine the intervals where \(f(x) > 0\) or \(f(x) < 0\), examine the sign of \(f(x)\) based on the critical points \(x = -3\) and \(x = 1\). These points divide the x-axis into intervals.1. Interval \((-\infty, -3)\): - Choose \(x = -4\): - \[f(-4) = ((-4 + 3)^4)((-4 - 1)^3) = 1^4 imes (-5)^3 = -125 < 0\]2. Interval \((-3, 1)\): - Choose \(x = 0\): - \[f(0) = (3)^4 imes (-1)^3 = -81 < 0\]3. Interval \((1, \infty)\): - Choose \(x = 2\): - \[f(2) = (5)^4 imes (1)^3 = 625 > 0\]Thus, \(f(x) < 0\) on \((-\infty, -3) \cup (-3, 1)\) and \(f(x) > 0\) on \((1, \infty)\).

Key Concepts

Y-Intercept

The y-intercept of a function is the point where the graph crosses the y-axis. This occurs when the value of the input variable \( x \) is zero. In simpler terms, it tells us what value \( y \) takes when \( x \) is zero. To find the y-intercept of the function \( f(x) = (x+3)^4(x-1)^3 \), we substitute \( x = 0 \) into the function. When we perform this calculation, we get:\[f(0) = (0 + 3)^4 \times (0 - 1)^3 = 3^4 \times (-1)^3 = 81 \times (-1) = -81\]Thus, the y-intercept is at the point \((0, -81)\). This means that the graph of the function will touch the y-axis at this point. Identifying the y-intercept is especially helpful in understanding how the graph behaves initially. If you're plotting a basic sketch, the y-intercept will give you a start point on the vertical axis.

X-Intercept

The x-intercepts are where a function graph crosses the x-axis. This happens when the output value \( y \) of the function is zero. Essentially, it's the value(s) of \( x \) when the entire equation equals zero. Let's determine the x-intercepts for our function:To find these intercepts in the function \( f(x) = (x+3)^4(x-1)^3 \), set the function equal to zero:\[ (x+3)^4(x-1)^3 = 0 \]Using the zero-product property (a principle stating that if a product is zero, at least one of the factors must be zero), we solve for each factor:- \((x+3)^4 = 0 \rightarrow x = -3\)- \((x-1)^3 = 0 \rightarrow x = 1\)Thus, the function's graph crosses the x-axis at two points, \((-3, 0)\) and \((1, 0)\). This means these are crucial points of interest where the function output changes sign, which plays an integral role in understanding the overall behavior of the function.

Intervals of Positivity and Negativity

Identifying the intervals of positivity and negativity of a function provides insights into where the function outputs are above or below zero. This knowledge helps to predict the behavior of the graph around certain ranges of \( x \). For the function \( f(x) = (x+3)^4(x-1)^3 \), the x-axis is split into different intervals by its critical points: \( x = -3 \) and \( x = 1 \).Consider these intervals:- **Interval \((-fty, -3)\):** Here, if we test a point like \( x = -4 \), it leads to \( f(-4) = -125 \), indicating the function is negative.- **Interval \((-3, 1)\):** Choosing \( x = 0 \) provides \( f(0) = -81 \), meaning the function remains negative in this range.- **Interval \((1, fty)\):** Testing \( x = 2 \), results in \( f(2) = 625 \), proving the function is positive.Hence, \( f(x) < 0 \) on the intervals \((-fty, -3) \cup (-3, 1)\) and \( f(x) > 0 \) on the interval \((1, fty)\). Recognizing these intervals aids comprehension of when the function's graph lies above or below the x-axis, providing critical information for comprehensive graph analysis.