Question

Assume that \(f\) is differentiable on \((-\infty, \infty)\) and \(x_{0}\) is a critical point of \(f\). (a) Let \(h(x)=f(x) g(x),\) where \(g\) is differentiable on \((-\infty, \infty)\) and $$ f\left(x_{0}\right) g^{\prime}\left(x_{0}\right) \neq 0 $$ Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) and the tangent line to the curve \(y=g(x)\) at \(\left(x_{0}, g\left(x_{0}\right)\right.\) intersect on the \(x\) -axis. (b) Suppose that \(f\left(x_{0}\right) \neq 0 .\) Let \(h(x)=f(x)\left(x-x_{1}\right),\) where \(x_{1}\) is arbitrary. Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the \(x\) -axis at \(\bar{x}=x_{1}\) (c) Suppose that \(f\left(x_{0}\right) \neq 0 .\) Let \(h(x)=f(x)\left(x-x_{1}\right)^{2},\) where \(x_{1} \neq x_{0} .\) Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the \(x\) -axis at the midpoint of the interval with endpoints \(x_{0}\) and \(x_{1}\). (d) Let \(h(x)=\left(a x^{2}+b x+c\right)\left(x-x_{1}\right),\) where \(a \neq 0\) and \(b^{2}-4 a c \neq 0 .\) Let \(x_{0}=-\frac{b}{2 a} .\) Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the \(x\) -axis at \(\bar{x}=x_{1}\). (e) Let \(h\) be a cubic polynomial with zeros \(\alpha, \beta,\) and \(\gamma,\) where \(\alpha\) and \(\beta\) are distinct and \(\gamma\) is real. Let \(x_{0}=\frac{\alpha+\beta}{2}\). Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the axis at \(\bar{x}=\gamma\)

Short answer

Answer: The tangent lines intersect the x-axis at the following points: (a) \(x=x_0\) (b) \(x=x_1\) (c) \(x=\frac{x_0 + x_1}{2}\) (d) \(x=x_1\) (e) \(x=\gamma\)

Step-by-step solution

Compute the derivative of h(x)

In this part, we want to find the tangent line to the curve \(y = h(x) = f(x)g(x)\) at the point \((x_0, h(x_0))\). To do this, we first need to find the derivative of h with respect to x, \(h'(x)\). Use the product rule for differentiation: \(h^\prime(x)=f^\prime(x)g(x)+f(x)g^\prime(x)\).

Determine the tangent line at \(x_0\)

The slope of the tangent line at \(x_0\) is given by the value of the derivative \(h'(x_0)\). Since \(x_0\) is a critical point of \(f\), \(f'(x_0)=0\). Therefore, \(h'(x_0)=0\cdot g(x_0)+f(x_0)g'(x_0)=f(x_0)g'(x_0)\). The equation of the tangent line to the curve \(y=h(x)\) at \(x_0\) can be expressed as: $$y-h(x_{0})=h'(x_0)(x-x_{0})$$ Now, we need to find the tangent line to the curve \(y = g(x)\) at \((x_0, g(x_0))\).

Find the equation of the tangent line to the curve \(y=g(x)\)

To find the tangent line at the point \((x_0, g(x_0))\), use the slope \(g'(x_0)\): $$y-g(x_{0})=g'(x_0)(x-x_{0})$$

Solve for the intersection point on the x-axis

Setting \(y=0\) in both tangent line equations, we get: >Tangent to \(y=h(x)\) at \((x_0, h(x_0))\): \(-h(x_0) = h'(x_0)(x - x_0)\) > >Tangent to \(y=g(x)\) at \((x_0, g(x_0))\): \(-g(x_0) = g'(x_0)(x - x_0)\) Observe that the intersection point on the x-axis occurs at \(x=x_{0}\) which satisfies both equations above given that \(f(x_0)g'(x_0) \neq 0\). --------- (b)

Compute h'(x)

Given \(h(x)=f(x)(x-x_1)\), with \(f'(x_0) = 0\), compute \(h'(x)\) using the product rule: \(h'(x) = f'(x)(x-x_1)+f(x)\).

Find the tangent line's equation and solve for the intersection point on the x-axis

The tangent line equation for the curve \(y=h(x)\) at \((x_0, h(x_0))\) is: $$y-h(x_0)=h'(x_0)(x-x_0)$$ Setting \(y=0\), we find the intersection point as \(x=x_1\). --------- (c)

Compute h'(x)

Given \(h(x)=f(x)(x-x_1)^{2}\) with \(f'(x_0) = 0\), compute \(h'(x)\) using the product rule: \(h'(x) = f'(x)(x-x_1)^2 + 2f(x)(x-x_1)\).

Find the tangent line's equation and solve for the intersection point on the x-axis

The tangent line equation for the curve \(y=h(x)\) at \((x_0, h(x_0))\) is: $$y - h(x_0) = h'(x_0)(x - x_0)$$ Setting \(y=0\), we find the intersection point as \(x = \frac{x_0 + x_1}{2}\), which is the midpoint of the interval with endpoints \(x_0\) and \(x_1\). --------- (d)

Compute h'(x)

Given \(h(x)=(ax^{2}+bx+c)(x-x_1)\) and \(x_0 = -\frac{b}{2a}\), compute \(h'(x)\) using the product rule: \(h'(x) = (2ax+b)(x-x_1)+(ax^{2}+bx+c)\).

Find the tangent line's equation and solve for the intersection point on the x-axis

The tangent line to the curve \(y = h(x)\) at \((x_0, h(x_0))\) is: $$y-h(x_{0})=h'(x_0)(x-x_{0})$$ Setting \(y=0\), we find the intersection point as \(x = x_1\). --------- (e)

Identify h'(x_0)

Given that \(h(x)\) is a cubic polynomial, we can express it in terms of its zeros: \(h(x) = k(x-\alpha)(x-\beta)(x-\gamma)\), with \(x_0 = \frac{\alpha + \beta}{2}\). Compute h'(x) and plug in the value of \(x_0\).

Find the tangent line's equation and solve for the intersection point on the x-axis

The tangent line to the curve \(y = h(x)\) at \((x_0, h(x_0))\) is: $$y-h(x_{0})=h'(x_0)(x-x_{0})$$ Setting \(y=0\), we find the intersection point as \(x=\gamma\).

Key concepts

Critical Points

In real analysis, a critical point of a differentiable function is a point on the graph where the derivative is zero or undefined. At critical points, the function's graph has a tangent that is horizontal or vertical, indicating a potential peak, trough, or saddle point. When a derivative is equal to zero, it suggests that the slope of the tangent line is zero, meaning the graph is flat at that point.
To determine the nature of the critical point—whether it represents a local maximum, minimum, or a saddle point—further examination is required. This can include analyzing the second derivative or using additional tests like the First or Second Derivative Test. Understanding critical points is essential in calculus as it helps in sketching the function and finding extreme values which are crucial in optimization problems.
Recognizing critical points involves:

• Identifying where the first derivative is zero or does not exist.
• Assessing the second derivative, if necessary, to ascertain the nature (maximum, minimum, inflection point).

Tangent Lines

The tangent line to a curve at a given point is the straight line that just "touches" the curve at that point. This line represents the immediate rate of change of the function at that specific location. If the curve was extended, the tangent line would be the best linear approximation to the curve at that point.
A key aspect of tangent lines is their slope, which is given by the derivative of the function at the point of tangency. When we find the equation of a tangent line, it often takes the form: \[ y - y_0 = m(x - x_0) \]
where \( m \) is the slope at the point \((x_0, y_0)\). This linear equation helps to visualize the function’s behavior locally. Knowing how to find and use tangent lines is paramount in applications such as physics and engineering, where they can represent instantaneous velocity or stress along an object.
To determine tangent lines, you generally:

• Find the derivative, which gives the slope of the tangent.
• Substitute the specific point into the derivative to find the slope at that point.

Differentiation

Differentiation is a fundamental mathematical process in calculus, used to determine the rate at which a quantity changes. The derivative, denoted as \( f'(x) \), is the result of differentiation and represents the slope of the tangent line to the function at any point \( x \). This makes it a powerful tool for analyzing how functions behave over intervals.
In the context of real analysis, differentiation provides insights into the behavior of complex functions, including the determination of critical points, concavity, and points of inflection. The procedures of differentiation follow specific rules, such as the power rule, chain rule, quotient rule, and product rule. Mastery of these rules is essential to solve real-world problems involving optimization and motion.
When differentiating, keep in mind:

• Always apply the appropriate rule based on the function's form.
• Check your work using derivative calculations to ensure accuracy.

Product Rule

The product rule is a fundamental technique used in differentiation when dealing with the derivative of a product of two or more functions. When you have a function that is a product of two functions, like \( h(x) = f(x)g(x) \), the product rule provides a method to differentiate it.
The formula for applying the product rule is:\[h'(x) = f'(x)g(x) + f(x)g'(x)\]This equation helps you to find the derivative of the product by taking the derivative of the first function, multiplying it by the second function, and adding the product of the first function with the derivative of the second function.
Understanding the product rule is crucial for accurately finding derivatives in complex mathematical modeling, physics, and engineering problems where functions frequently multiply. It ensures that all components of the function product are considered during differentiation.
When using the product rule, it is important to:

• Identify both functions in the product correctly.
• Compute the derivatives of each function separately before applying the rule.