Problem 44 - Calculus for Biology and Medicine

Question

In Problems , find the tangent line, in standard form, to \(y=\) \(f(x)\) at the indicated point. $$ y=-2 x^{3}-3 x+1, \text { at } x=1 $$

Step-by-Step Solution

Answer

The tangent line in standard form is \(9x + y = 5\).

Step 1: Differentiate the Function

To find the equation of the tangent line, we need the slope of the tangent which is the derivative of the function. Differentiate the function \(y = -2x^3 - 3x + 1\) with respect to \(x\): \[ f'(x) = \frac{d}{dx}(-2x^3 - 3x + 1) = -6x^2 - 3. \]

Step 2: Evaluate the Derivative at the Given Point

We need to find the slope of the tangent at \(x = 1\). Substitute \(x = 1\) into the derivative: \[ f'(1) = -6(1)^2 - 3 = -6 - 3 = -9. \] The slope of the tangent line at \(x = 1\) is \(-9\).

Step 3: Find the Point on the Curve

The tangent line passes through the point on the curve at \(x = 1\). Find \(y\) when \(x = 1\): \[ y = -2(1)^3 - 3(1) + 1 = -2 - 3 + 1 = -4. \] The point on the curve is \((1, -4)\).

Step 4: Apply the Point-Slope Formula

Use the point-slope form to find the equation of the tangent line: \[ y - y_1 = m(x - x_1), \] where \((x_1, y_1) = (1, -4)\) and \(m = -9\). So: \[ y + 4 = -9(x - 1). \] Simplify to: \[ y + 4 = -9x + 9. \] \[ y = -9x + 5. \]

Step 5: Convert to Standard Form

Convert \(y = -9x + 5\) to the standard form \(Ax + By = C\): Add \(9x\) to both sides:\[ 9x + y = 5. \] The standard form equation of the tangent line is \(9x + y = 5\).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a function is changing. In simpler terms, it tells us how steep a curve is at any given point. When you have a function, like our cubic function \( y = -2x^3 - 3x + 1 \), its derivative will give you a new function, which reveals the slope of the original function at every point along its curve. For our problem, differentiating the function with respect to \( x \) gives \( f'(x) = -6x^2 - 3 \).

Here's how you differentiate:

• Each term is processed separately. For a term \( ax^n \), its derivative is \( nax^{n-1} \).
• In our function, \(-2x^3\) becomes \(-6x^2\) (since \(3 \times -2 = -6\)), and \(-3x\) becomes \(-3\), while constants like \(+1\) disappear.

Once you have your derivative, you can find the slope of the tangent line by plugging in the value of \(x\) at the point of interest. Here, substituting \(x = 1\) into \(-6x^2 - 3\) gave us \(-9\), which is the slope at that point.

Point-Slope Formula

The point-slope formula offers a straightforward way to find the equation of a line when you know a point it passes through and its slope. This form is expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a known point on the line, and \( m \) is the slope. In our context:

• The point \((1, -4)\) is derived from evaluating your original function at \(x = 1\).
• The slope \(-9\) is determined from the derivative at the same \(x\) value.

Substituting these into the formula gives us:

\( y + 4 = -9(x - 1) \).

This equation captures the line that just "touches" the curve at the point \( (1, -4) \), perfectly representing the tangent.

Standard Form

Standard form is a way of expressing the equation of a line with integer coefficients on one side of the equation, typically as \(Ax + By = C\). This form is particularly useful in various mathematical applications because it provides a clear and balanced way to represent linear equations.

To convert our tangent line from point-slope to standard form:

• Start with your point-slope equation, \(y + 4 = -9(x - 1)\).
• Simplify it to the slope-intercept form, \(y = -9x + 5\).
• Add \(9x\) to both sides to shift terms: \(9x + y = 5\).

This transformation gives us an equation \(9x + y = 5\) that is neatly arranged in standard form. It's crucial to move terms around to have \(x\) and \(y\) on one side and the constant on the other, ensuring the equation is balanced and easier to interpret for certain mathematical operations.