/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Assume \(f\) is a differentiable... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 80

Assume \(f\) is a differentiable function whose graph passes through the point \((1,4) .\) Suppose \(g(x)=f\left(x^{2}\right)\) and the line tangent to the graph of \(f\) at (1,4) is \(y=3 x+1 .\) Determine each of the following. a. \(g(1)\) b. \(g^{\prime}(x)\) c. \(g^{\prime}(1)\) d. An equation of the line tangent to the graph of \(g\) when \(x=1\)

Short Answer

Expert verified
2. What is \(g'(x)\)? 3. What is \(g'(1)\)? 4. What is the equation of the tangent line to the graph of \(g\) at \(x = 1\)?

Step by step solution

01

Find \(g(1)\).

To find \(g(1)\), we need to replace \(x\) with \(1\) in the formula for \(g(x)\): $$ g(x) = f\left(x^2\right) \Rightarrow g(1) = f\left(1^2\right) = f(1). $$ Since we know that the point \((1,4)\) is on the graph of \(f\), we can infer that \(f(1) = 4\). Therefore, \(g(1) = 4\).
02

Find \(g'(x)\).

To find the derivative of \(g(x)\) with respect to \(x\), we will use the Chain Rule: $$ g'(x) = \frac{d}{dx}\left[f\left(x^2\right)\right] = f'\left(x^2\right) \cdot \frac{d}{dx}\left[x^2\right]. $$ We know that the derivative of \(x^2\) with respect to \(x\) is \(2x\) (by the Power Rule). Thus, we have: $$ g'(x) = f'\left(x^2\right) \cdot 2x. $$
03

Find \(g'(1)\).

To find \(g'(1)\), we need to replace \(x\) with \(1\) in the formula for \(g'(x)\): $$ g'(1) = f'\left(1^2\right) \cdot 2\left(1\right) = f'(1) \cdot 2. $$ Since the tangent line to the graph of \(f\) is given by \(y = 3x + 1\), we know that the slope of the tangent line at point \((1,4)\), which is equal to \(f'(1)\), is 3. Therefore, \(g'(1) = 3 \cdot 2 = 6\).
04

Find the equation of the tangent line to the graph of \(g\) at \(x=1\).

We know that \(g(1) = 4\) and \(g'(1) = 6\). Thus, the point \((1,4)\) is on the graph of \(g\) and the slope of the tangent line at this point is 6. Using the point-slope form, we can write the equation of the tangent line as: $$ y - 4 = 6(x - 1) \Rightarrow y = 6x - 2. $$ So, the equation of the tangent line to the graph of \(g\) when \(x=1\) is \(y = 6x - 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line
The tangent line to a function at a given point is a straight line that touches the graph of the function at that point, without crossing it. It gives us an approximation of the function near that point. The slope of this tangent line represents the derivative of the function at that specific point.

In the given exercise, the equation of the tangent line to the function \(f\) at the point \((1,4)\) is \(y = 3x + 1\). This indicates that the slope of the tangent line, or \(f'(1)\), is 3. This slope tells us how steeply the function \(f\) is rising as \(x\) approaches 1.
  • The point of tangency is where the line and the graph meet, which here is \((1,4)\).
  • Understanding the tangent line's function is critical for solving differential calculus problems as it provides a linear approximation.
Chain Rule
The chain rule is a fundamental principle in calculus used to find the derivative of a composite function. When we have a function inside another function, the chain rule helps us differentiate it. The formula for the chain rule is:

\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

In the exercise, we need to find \(g'(x)\) where \(g(x) = f(x^2)\). The chain rule is applied to take the derivative of \(g\) as a composition of \(f\) and \(x^2\). We let \(u = x^2\), so \(g(x) = f(u)\). Following the chain rule:
  • \(g'(x) = f'(u) \cdot \frac{du}{dx}\)
  • \(\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x\)
So, the derivative \(g'(x) = f'(x^2) \cdot 2x\). The chain rule makes it possible to handle complex functions systematically and accurately.
Power Rule
The power rule is one of the most straightforward yet essential differentiation rules. It is used to find the derivative of any function of the form \(x^n\), where \(n\) is any real number. The power rule states that:

If \(y = x^n\), then \(\frac{dy}{dx} = nx^{n-1}\).

In our problem, to find the derivative of \(x^2\) in \(g(x) = f(x^2)\), we use the power rule:
  • For \(x^2\), \(n = 2\), so \(\frac{d}{dx}(x^2) = 2x^{2-1} = 2x\).
This rule simplifies the entire process of finding derivatives, especially when combined with other rules, such as the chain rule, in more complex problems like the one presented. Mastery of the power rule is crucial for tackling calculus effectively.
Derivative Calculation
Derivative calculation is at the heart of calculus and provides us with information on how a function changes at any point. In this exercise, calculating the derivatives involves understanding both the power rule and the chain rule to handle composite functions like \(g(x) = f(x^2)\).

Here鈥檚 a quick overview of the steps taken in the solution:
  • Recognize the need for the chain rule due to the composition \(g(x) = f(x^2)\).
  • Use the power rule to find the derivative of the inner function \(x^2\), which is \(2x\).
  • Apply the chain rule: \(g'(x) = f'(x^2) \cdot 2x\).
With regard to the point \(x = 1\), it鈥檚 given that the slope of the tangent to \(f\) at \((1,4)\) is 3, thus \(f'(1) = 3\). Substituting in, we find \(g'(1) = 6\), indicating a steepness or rate of change for \(g\) at that point. Derivative calculations help us understand the dynamic behavior of functions.

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Most popular questions from this chapter

Tangent lines and exponentials. Assume \(b\) is given with \(b>0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, 28, Mar 1997).

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \sqrt{10 x}$$,
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