Question

Derivatives and tangent lines a. For the following functions and values of \(a,\) find \(f^{\prime}(a)\) b. Determine an equation of the line tangent to the graph of \(f\) at the point \((a, f(a))\) for the given value of \(a\) $$f(x)=2 x^{3} ; a=10$$

Short answer

Answer: The equation of the tangent line at the point where \(a=10\) is \(y = 600x - 4000\).

Step-by-step solution

Find the derivative of the function

To find the derivative of the function \(f(x) = 2x^3\), we'll use the power rule, which states that the derivative of \(x^n\) concerning \(x\) is \(nx^{n-1}\). So, for \(f(x) = 2x^3\), the derivative is \(f'(x) = 6x^2\).

Evaluate the derivative at \(a=10\)

Now, we need to find the value of the derivative at \(a=10\). To do this, we'll substitute \(x=10\) into the derivative function: $$f'(10) = 6(10)^2 = 6\cdot100 = 600$$ So, the value of the derivative at \(a=10\) is 600.

Find the value of the function at \(a=10\)

To find the coordinates of the point on the graph where the tangent line touches, we need to find the value of the function at \(a=10\). We'll substitute \(x=10\) into the original function: $$f(10) = 2(10)^3 = 2\cdot1000 = 2000$$ So, the point on the graph at \(a=10\) is \((10, 2000)\).

Determine the equation of the tangent line

To find the equation of the tangent line, we'll use the point-slope form of a linear equation, which is given by: $$y-y_1 = m(x-x_1)$$ where \((x_1, y_1)\) is the point on the graph, and \(m\) is the slope of the tangent line (which is the value of the derivative at \(a=10\)). Plugging in the values: $$y - 2000 = 600(x - 10)$$ Now, we'll simplify the equation to get it into slope-intercept form (\(y = mx + b\)): $$y = 600x - 600\cdot10 + 2000$$ $$y = 600x - 4000$$ So, the equation of the tangent line is: $$y = 600x - 4000$$

Key concepts

Derivatives

In calculus, derivatives measure how a function changes as its input changes. Imagine that a function represents a moving car's distance over time. Its derivative tells us the car's speed at each instant. Derivatives are vital because they provide insights about function behavior such as increasing or decreasing trends.

Finding the derivative of a function involves mathematical rules and techniques. One popular method is the power rule, which significantly simplifies the process for polynomials. Derivatives aren't only important for analyzing functions on their own but also serve crucial roles in determining tangent lines and optimizing problems.

Overall, derivatives help us understand the underlying sentiment of a graph or function, guiding predictions and calculations related to real-world phenomena.

Tangent Lines

A tangent line is like a snapshot of a curve at a specific point, providing a linear approximation of the curve's behavior at that point. In practical terms, the tangent line represents the best line that "just touches" the curve without crossing it. By knowing where the tangent line touches the curve, we gain valuable information about the curve's direction and steepness at that specific point.

To find the equation of a tangent line, we need two things:

• The point of tangency, where the curve and tangent line meet, \(a, f(a)\).
• The slope of the tangent line, given by the derivative of the function at the point of tangency (\(f'(a)\)).

With these, we apply the point-slope form of a line equation to express the tangent line mathematically. Tangent lines are critical in calculus as they simplify complex curves into manageable components, aiding in analysis and predictions.

Power Rule

The power rule is a fundamental technique in calculus for finding derivatives of polynomial expressions. It provides a quick and straightforward way to determine the rate of change of functions of the form \(x^n\). The basic idea is simple: take the exponent from \(x^n\), multiply it by the coefficient, and reduce the exponent by one. This rule turns lengthy derivative calculations into a simple two-step process.

For instance, if you have \(f(x) = 2x^3\), applying the power rule means:

• Take the exponent \(3\), multiply it by the coefficient \(2\), giving \(3 imes 2 = 6\).
• Lower the exponent by \(1\), changing it from \(3\) to \(2\).

Thus, the derivative is \(f'(x) = 6x^2\).

The power rule applies to any term with an \(x^n\) format, making it invaluable for quickly finding derivatives of many standard algebraic functions. It's one of the first rules students learn in calculus classes due to its ease and powerful results.