Question

Assume \(f\) and \(g\) are differentiable on their domains with \(h(x)=f(g(x)) .\) Suppose the equation of the line tangent to the graph of \(g\) at the point (4,7) is \(y=3 x-5\) and the equation of the line tangent to the graph of \(f\) at (7,9) is \(y=-2 x+23\) a. Calculate \(h(4)\) and \(h^{\prime}(4)\) b. Determine an equation of the line tangent to the graph of \(h\) at \((4, h(4))\)

Short answer

Answer: h(4) = 9, h'(4) = -6, and the equation of the tangent line is \(y = -6x + 33\).

Step-by-step solution

Calculate h(4)

h(x) = f(g(x)). Since the tangent line to the graph of g at the point (4, 7) is given by \(y = 3x - 5\), we know that g(4) = 7. Using this information, we can find h(4): h(4) = f(g(4)) h(4) = f(7) Now, we know the tangent line to the graph of f at the point (7, 9) is given by \(y=-2x+23\). This means that f(7) = 9. Therefore, h(4) = 9.

Find g'(4) using the tangent line to g at (4, 7)

The equation of the tangent line to the graph of g at (4, 7) is \(y = 3x - 5\). The slope of this line is 3, which represents the derivative of g at x = 4. So, g'(4) = 3.

Find f'(7) using the tangent line to f at (7, 9)

The equation of the tangent line to the graph of f at (7, 9) is \(y = -2x + 23\). The slope of this line is -2, which represents the derivative of f at x = 7. So, f'(7) = -2.

Calculate h'(4) using the chain rule

To find h'(4), we will use the chain rule, which states that \({\displaystyle h'(x)=f'(g(x))g'(x)}\). In our case, we need to evaluate this at x = 4: h'(4) = f'(g(4)) * g'(4) h'(4) = f'(7) * 3 h'(4) = -2 * 3 h'(4) = -6 Thus, h(4) = 9 and h'(4) = -6. #b. Determine an equation of the line tangent to the graph of \(h\) at $(4, h(4))$#

Use the point-slope form of a linear equation

The equation of the line tangent to the graph of h at the point $(4, h(4))\( can be found using the point-slope form of a linear equation, which is \)y - y_1 = m(x - x_1)$. We know the point (4, h(4)) = (4, 9) and the slope, h'(4) = -6. Using this information, we can write the equation of the tangent line: \(y - 9 = -6(x - 4)\)

Rewriting the equation in a more simplified form (y = mx + b)

Let's rewrite the equation in slope-intercept form (y = mx + b): \(y - 9 = -6(x - 4)\) \(y = -6x + 24 + 9\) \(y = -6x + 33\) The equation of the line tangent to the graph of h at the point (4, h(4)) is \(y = -6x + 33\).

Key concepts

Tangent Line Equation

In calculus, the tangent line represents the instantaneous direction of a function at a particular point. It essentially gives us the best linear approximation of the function around a specific input value. To find the equation of a tangent line, we commonly use the point-slope form of a linear equation, which is expressed as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the tangent line and \(x_1\), \(y_1\) are the coordinates of the point of tangency.

Calculating the equation of a tangent line to the curve \(y = f(x)\) at a given point \(x = a\) involves two steps. First, determine \(f'(a)\), the derivative of the function at \(x = a\), which gives the slope \(m\) of the tangent. Second, use the coordinates of the point of tangency \(a, f(a)\) to solve for the equation of the tangent line using the point-slope formula.

Derivative Calculation

The derivative of a function at any point captures the rate at which the function's value changes with respect to a change in its input value. It is a fundamental concept in calculus, used to determine instantaneous rates of change, slopes of tangent lines, and more. The calculation of a derivative depends on the function type – power, exponential, trigonometric, and others all have specific rules for differentiation.

The process for deriving the function often employs rules such as the power rule, the chain rule, the product rule, and the quotient rule. The case we encountered in the exercise used the chain rule for composite functions, which states that if \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\). In our example, the derivative calculation required understanding the tangent line (which represents the derivative at a specific point) for each individual function before applying the chain rule to find the composite function’s derivative at a point.

Composite Function Differentiation

Differentiation of composite functions is a process that involves the application of the chain rule, a pivotal tool in calculus. The chain rule deals with the derivative of a function composed of two or more functions, stating that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.

For instance, if we have two functions \(f(x)\) and \(g(x)\), and their composite is \(h(x) = f(g(x))\), the chain rule tells us that \(h'(x) = f'(g(x)) \cdot g'(x)\). Implementing this rule requires careful attention to correctly apply the derivatives of the inner and outer functions. The exercise provided a practical overview of employing the chain rule. It involved first finding the derivatives of the inner function \(g\) and the outer function \(f\) based on their tangent lines. Then we multiplied these derivatives to obtain \(h'(x)\), the derivative of the composite function \(h\) at a given point. This process is key when dealing with complex functions and is a staple in advanced calculus problems.