Question

Suppose \(f(3)=1\) and \(f^{\prime}(3)=4 .\) Let \(g(x)=x^{2}+f(x)\) and \(h(x)=3 f(x)\) a. Find an equation of the line tangent to \(y=g(x)\) at \(x=3\) b. Find an equation of the line tangent to \(y=h(x)\) at \(x=3\)

Short answer

Answer: The equation of the tangent line to \(y=g(x)\) at \(x=3\) is \(y=10x-20\), and the equation of the tangent line to \(y=h(x)\) at \(x=3\) is \(y=12x-33\).

Step-by-step solution

Find the derivative of \(g(x)\)

To find the derivative of \(g(x)=x^{2}+f(x)\), we can use the sum rule for differentiation: the derivative of the sum of two functions is the sum of their derivatives. So we need to find the derivatives of \(x^{2}\) and \(f(x)\) separately. We know that \(\frac{d}{dx}(x^2) = 2x\), and we are given that \(f^{\prime}(3)=4\). The derivative of \(g(x)\) is given by \(g^{\prime}(x) = 2x + f^{\prime}(x)\).

Use the derivative of \(g(x)\) to find the tangent line at \(x=3\)

We know that \(g(3)=3^2+f(3)=9+1=10\), and the slope of the tangent line at the point \((3,10)\) is given by the value of \(g^{\prime}(3)=2(3)+4=10\). Using the point-slope form of a linear equation, we get the equation for the tangent line to \(g(x)\) at \(x=3\): \(y-10=10(x-3)\), which simplifies to \(y=10x-20\). a. The equation of the line tangent to \(y=g(x)\) at \(x=3\) is \(y=10x-20\).

Find the derivative of \(h(x)\)

Now we need to find the derivative of \(h(x)=3f(x)\). Using the constant multiple rule for differentiation, we get \(h^{\prime}(x)=3f^{\prime}(x)\).

Use the derivative of \(h(x)\) to find the tangent line at \(x=3\)

We know that \(h(3)=3f(3)=3(1)=3\), and the slope of the tangent line at the point \((3,3)\) is given by the value of \(h^{\prime}(3)=3f^{\prime}(3)=3(4)=12\). Using the point-slope form of a linear equation, we get the equation for the tangent line to \(h(x)\) at \(x=3\): \(y-3=12(x-3)\), which simplifies to \(y=12x-33\). b. The equation of the line tangent to \(y=h(x)\) at \(x=3\) is \(y=12x-33\).

Key concepts

Differentiation Rules

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of the function with respect to one of its variables. Several rules exist for finding derivatives, simplifying the process for different types of functions, such as polynomial, exponential, and trigonometric functions.

The sum rule in differentiation is straightforward. It states that the derivative of a sum of two functions is the sum of their individual derivatives. For instance, if you have two functions, say, \(u(x)\) and \(v(x)\), then the derivative, \(\frac{d}{dx}(u(x) + v(x))\), can be written as \(\frac{d}{dx} u(x) + \frac{d}{dx} v(x)\). This was applied in the solution to find the derivative of \(g(x) = x^2 + f(x)\), resulting in \(2x + f^{\prime}(x)\).

Another useful rule is the constant multiple rule. This rule simplifies the process when a function is multiplied by a constant. The derivative of \(c \cdot u(x)\), where \(c\) is some constant, is simply \(c \cdot \frac{d}{dx} u(x)\). In the exercise, we used this rule for \(h(x) = 3f(x)\), deriving \(h^{\prime}(x) = 3f^{\prime}(x)\).

Point-Slope Form

The point-slope form is a method used to find the equation of a straight line in coordinate geometry. It is particularly handy when you know one point on the line and the slope. The formula can be expressed as:

• \(y - y_1 = m(x - x_1)\)

Here, \((x_1, y_1)\) defines the known point and \(m\) is the slope of the line. This form is incredibly useful for writing equations of tangent lines, which touch a curve at exactly one point and share the same slope as the curve at that point.

In the solution provided for both \(g(x)\) and \(h(x)\), we used the point-slope form to determine the tangent line at \(x = 3\). For \(g(x)\), the slope was calculated as \(10\), giving us the equation \(y - 10 = 10(x - 3)\), which simplifies to \(y = 10x - 20\). For \(h(x)\), with a slope of \(12\), the equation obtained was \(y - 3 = 12(x - 3)\), simplified to \(y = 12x - 33\).

Derivative Calculation

Calculating derivatives is central in calculus as it helps describe how functions behave. Derivatives can represent different aspects, such as slopes of tangent lines or rates of change. In the exercise, we calculated derivatives using differentiation rules to solve for tangent lines.

To find the derivative of \(g(x) = x^2 + f(x)\), the sum rule was used, giving \(g^{\prime}(x) = 2x + f^{\prime}(x)\). Specific values were then substituted into the derivative to determine the slope at a particular point. Here, \(g^{\prime}(3) = 2(3) + 4 = 10\).

For \(h(x) = 3f(x)\), the derivative was found using the constant multiple rule, so \(h^{\prime}(x) = 3f^{\prime}(x)\). At \(x = 3\), this performed as \(h^{\prime}(3) = 3 \cdot 4 = 12\).

• These derivatives illustrate the slopes of the tangent lines for \(g(x)\) and \(h(x)\) at \(x = 3\), essential for finding their equations.

Simplifying derivatives involves applying these rules correctly and often makes solving complex problems much easier.