Problem 74 - Calculus for Biology and Medicine

Question

Find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. \(f(x)=\sqrt{x}\left(x^{3}-1\right)\), at \(x=1\)

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 3x - 3\).

Step 1: Find the function value at the point

Compute the value of the function at the point where the tangent line is needed, which is at \(x = 1\). Evaluate \(f(1) = \sqrt{1}(1^3 - 1) = 1(1-1) = 0\). So, the point is \((1, 0)\).

Step 2: Calculate the derivative of the function

First find \(f(x) = \sqrt{x}(x^3 - 1)\). To find \(f'(x)\), apply the product rule: \(u = \sqrt{x} = x^{1/2}\) and \(v = x^3 - 1\). The derivatives are \(u' = \frac{1}{2}x^{-1/2}\) and \(v' = 3x^2\). The product rule states: \((uv)' = u'v + uv'\). So, \(f'(x) = \frac{1}{2}x^{-1/2}(x^3 - 1) + x^{1/2}(3x^2)\).

Step 3: Simplify the derivative

Simplify the expression for \(f'(x)\) obtained in Step 2. Expand and combine: \(f'(x) = \frac{1}{2}x^{-1/2}(x^3 - 1) + 3x^{2.5}\). This simplifies to: \(f'(x) = \frac{1}{2}x^{5/2} - \frac{1}{2}x^{-1/2} + 3x^{2.5}\). Combine like terms to simplify further.

Step 4: Evaluate the derivative at the point

Evaluate \(f'(x)\) at \(x=1\): \(f'(1) = \frac{1}{2}(1)^{5/2} - \frac{1}{2}(1)^{-1/2} + 3(1)^{2.5}\). This results in \(f'(1) = \frac{1}{2} - \frac{1}{2} + 3 = 3\). Thus, the slope \(m\) is 3.

Step 5: Form the equation of the tangent line

Use the point-slope form of a line to find the tangent line equation: \(y - y_1 = m(x - x_1)\). With the slope \(m = 3\) and point \((1, 0)\), we have \(y - 0 = 3(x - 1)\). Simplify to get \(y = 3x - 3\), the equation of the tangent line in slope-intercept form.

Key Concepts

Tangent Line

The tangent line to a curve at a specific point gives us the best linear approximation of the function near that point. Imagine the tangent line as just barely touching the curve at precisely one point, offering an exact straight-line representation of the curve's slope at that instant.

• In the provided exercise, the function is given as \(f(x) = \sqrt{x} (x^3 - 1)\). We find the tangent line at the point \(x = 1\).
• The function value is \(f(1) = 0\), meaning the tangent line touches the curve at the point \((1, 0)\).
• The tangent line equation is found using point-slope form: \(y - y_1 = m(x - x_1)\). For this problem, the slope \(m\) is found to be 3, resulting in the equation \(y = 3x - 3\).

This line touches the curve at one specific point, giving a brief glance at the curve's behavior around \(x = 1\). This small interaction is powerful, enabling us to understand the instantaneous rate of change.

Derivative

Derivatives are fundamental to calculus because they indicate how a function changes. A derivative at a particular point on a curve provides the slope of the tangent line to the curve at that point.

• To derive \(f(x) = \sqrt{x} (x^3 - 1)\), we need to calculate \(f'(x)\). The derivative reveals how fast or slow the function's value is changing as we move along the x-axis.
• In solving the exercise, we calculate \(f'(1)\) to determine the slope of the tangent line at \(x = 1\).
• The computed derivative \(f'(1) = 3\) implies that at \(x = 1\), the function is steeply climbing upwards, which means there's a positive rate of change at this point.

Thus, the derivative provides the key clue to form the tangent equation, illustrating how calculus translates dynamic function behavior into a static, linear prediction.

Product Rule

The product rule is an essential tool in calculus for differentiating products of two functions. When you have a function that is the product of two expressions, like \(u(x)\) and \(v(x)\), the product rule allows you to find its derivative accurately.

• The product rule formula is \((uv)' = u'v + uv'\), which means to differentiate the first times the second plus the second times the derivative of the first.
• In our exercise, with \(u = \sqrt{x} = x^{1/2}\) and \(v = x^3 - 1\), we apply this rule. Each function is differentiated separately, producing \(u' = \frac{1}{2}x^{-1/2}\) and \(v' = 3x^2\).
• After applying the product rule, we get \(f'(x) = \frac{1}{2}x^{-1/2}(x^3 - 1) + x^{1/2}(3x^2)\).

Using this rule simplifies finding the derivatives of complex multiplied functions, allowing you to accurately determine behaviors of compound expressions. It's a calculus engineer's way of tackling multifaceted dynamic function scenarios.