Problem 25 - Calculus for Biology and Medicine

Question

(a) Graph \(f(x)=x^{2}\) and \(g(x)=x^{3}\) for \(x \geq 0\), together, in one coordinate system. (b) Show algebraically that \(x \geq x^{2}\) for \(0 \leq x \leq 1\). (c) Show algebraically that \(x \leq x^{2}\) for \(x \geq 1\).

Step-by-Step Solution

Answer

(a) Graph both functions on the same axis. (b) Shows \(x \geq x^2\) for \(0 \leq x \leq 1\). (c) Shows \(x \leq x^2\) for \(x \geq 1\).

Step 1: Understanding the Functions

The functions are given as \(f(x) = x^2\) and \(g(x) = x^3\). We need to graph these functions together for \(x \geq 0\).

Step 2: Graphing the Functions

For \(x \geq 0\), plot both \(f(x) = x^2\) and \(g(x) = x^3\) on the same coordinate system. The parabola \(f(x) = x^2\) starts at the origin and curves upwards, while the cubic function \(g(x) = x^3\) also starts at the origin but increases more steeply as \(x\) increases.

Step 3: Analyzing the Inequality \(x \geq x^2\) for \(0 \leq x \leq 1\)

Consider the inequality \(x \geq x^2\). This can be rewritten as \(x - x^2 \geq 0\). Factoring gives us \(x(1-x) \geq 0\). Since \(x \geq 0\) and \(1 - x \geq 0\) for \(x \leq 1\), the inequality holds true for \(0 \leq x \leq 1\).

Step 4: Analyzing the Inequality \(x \leq x^2\) for \(x \geq 1\)

Consider the inequality \(x \leq x^2\). This becomes \(x^2 - x \geq 0\). Factoring gives \(x(x - 1) \geq 0\). Since \(x \geq 1\), both \(x \geq 0\) and \(x - 1 \geq 0\), thus confirming the inequality for \(x \geq 1\).

Key Concepts

Graphing functions

When graphing functions like \(f(x) = x^2\) and \(g(x) = x^3\), it's important to first understand what each function represents. The function \(f(x) = x^2\) is a quadratic function, resulting in a parabola. It touches the origin and curves upwards as \(x\) increases. In contrast, the function \(g(x) = x^3\) is a cubic function, which also starts at the origin but grows more steeply than the parabola due to the cubic term.Graphing them together for \(x \geq 0\), we can observe that:

• Both functions pass through the point (0,0).
• For values of \(0 \leq x \leq 1\), \(g(x)\) grows slower than \(f(x)\) because \(x^3\) increases more gently in this range.
• As \(x\) grows larger beyond 1, \(g(x)\) overtakes \(f(x)\), as the cubic growth outpaces the quadratic.

Seeing these functions plotted together offers a great visual understanding of how each grows over different intervals.

Inequality analysis

Analyzing inequalities for different intervals helps us understand how the mathematical relationships compare between functions. Let's consider the inequalities given here.First, the inequality \(x \geq x^2\) for \(0 \leq x \leq 1\):

• Rewrite the inequality as \(x - x^2 \geq 0\).
• This can be factored as \(x(1-x) \geq 0\).
• Since \(0 \leq x \leq 1\), both components \(x\) and \(1-x\) are non-negative.
• Hence, the product \(x(1-x)\) is non-negative, confirming the inequality holds for this range.

Next, consider \(x \leq x^2\) for \(x \geq 1\):

• Here, the inequality can be rewritten as \(x^2 - x \geq 0\).
• Factoring results in \(x(x-1) \geq 0\).
• For \(x \geq 1\), both \(x\) and \(x-1\) are positive, making the product non-negative.
• Thus, the inequality \(x \leq x^2\) is satisfied for \(x \geq 1\).

These analyses reveal how the relationships between the functions change over different intervals of \(x\).

Algebraic manipulation

Algebraic manipulation is the process of transforming equations or inequalities to simplify or solve them. In the given problems, this involves factoring and rearranging terms to make understanding easier.For instance, take the inequality \(x \geq x^2\) for \(0 \leq x \leq 1\):

• Start by rearranging to \(x - x^2 \geq 0\).
• The expression \(x - x^2\) can be rewritten by factoring out \(x\), giving \(x(1-x) \geq 0\).
• Such factoring makes it easy to check the non-negativity of each factor for the specified range.

Similarly, for \(x \leq x^2\) when \(x \geq 1\):

• Rearrange it to \(x^2 - x \geq 0\).
• Factor as \(x(x-1) \geq 0\).
• This simple factoring shows quickly that the inequality holds, given the non-negative values for \(x\) in this range.

Algebraic manipulation like factoring helps solve inequalities by breaking down complex expressions into simpler, more manageable forms.