Problem 22 - Calculus Early Transcendentals

Question

If $$g(\mathrm{x})=1-\mathrm{x}^{3},\( find \)g^{\prime}(0)$$ and use it to find an equation of the tangent line to the curve $$\mathrm{y}=1-\mathrm{x}^{3}$$ at the point \((0,1)\) .

Step-by-Step Solution

Answer

The tangent line equation is \( y = 1 \).

Step 1: Find the Derivative

To find the derivative \( g'(x) \), apply the power rule to the function \( g(x) = 1 - x^3 \). The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). This gives us \( g'(x) = 0 - 3x^2 = -3x^2 \).

Step 2: Evaluate the Derivative at x = 0

Substitute \( x = 0 \) into the derivative \( g'(x) = -3x^2 \) to find \( g'(0) \). This results in \( g'(0) = -3(0)^2 = 0 \). So, \( g'(0) = 0 \).

Step 3: Equation of the Tangent Line

The equation of the tangent line to the curve at a given point \((x_0, y_0)\) is \( y - y_0 = m(x - x_0) \), where \( m = g'(x_0) \) is the slope of the tangent. At the point \((0, 1)\), the slope \( m \) is \( 0 \) (since \( g'(0) = 0 \)). Therefore, the equation becomes \( y - 1 = 0(x - 0) \), which simplifies to \( y = 1 \).

Key Concepts

Power Rule

Understanding the power rule makes finding derivatives much simpler. It's a fundamental tool in calculus used to differentiate functions of the form \(x^n\), where \(n\) is any real number. The rule states that the derivative of \(x^n\) is \(nx^{n-1}\). This means you take the exponent \(n\), bring it in front of the \(x\), and then subtract one from the original exponent.Let's apply this power rule to the function given in the exercise: \(g(x) = 1 - x^3\).

• First, let's focus on the term \(-x^3\). According to the power rule, its derivative is \(-3x^{2}\) (we brought down the 3, and decreased the power by one).
• The derivative of a constant, like 1, is always 0, because constants do not change.

Thus, the derivative \(g'(x) = -3x^2\). This derivative tells us how the original function \(g(x)\) changes at any point \(x\). It is particularly useful for finding the slope of a curve at specific points.

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point. At this point, the tangent line has the same slope as the curve. This line is very useful to understand how the curve behaves at that point.To find the equation of the tangent line, remember this formula: \(y - y_0 = m(x-x_0)\). Here, \((x_0, y_0)\) is the point where the tangent line touches the curve and \(m\) is the slope of the tangent line at this point.

• For the given function, we identify the point of tangency as \((0, 1)\).
• In our exercise, we already found that \(g'(0) = 0\). This tells us the slope \(m\) at the point \((0, 1)\) is zero.

Plug in these values into the tangent line formula to get \(y - 1 = 0(x - 0)\). Simplifying this gives us the equation \(y = 1\). This means the tangent line at this point is horizontal, showing no change in \(y\) as \(x\) varies at \((0, 1)\).

Slope of a Curve

The slope of a curve at a specific point gives you an instantaneous rate of change. When computing it through derivatives, we determine how steep the curve is at that point.In calculus, the derivative itself represents a way to find this slope. For the function \(g(x)\), we already determined its derivative is \(-3x^2\). This expression tells us the slope of \(g(x)\) at any given \(x\)-value.Let's evaluate the slope at \(x = 0\):

• Substitute \(x=0\) into the derivative \(-3x^2\), which gives \(g'(0) = -3(0)^2 = 0\).
• This indicates that at \(x=0\), the curve is completely flat or horizontal. In this case, the slope of the curve coincides with the slope of the tangent line.

This kind of analysis helps you visualize how the function behaves locally—meaning right around the point you're evaluating—giving insights into the nature of the curve's motion and direction.