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Chapter 3: Problem 72

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

Short Answer

Expert verified
The equation of the tangent line to y=g(x) at x=2 is y - 8 = 20(x - 2), and the equation of the tangent line to y=h(x) at x=2 is y + 2 = -3(x - 2).

Step by step solution

01

Find g'(x)

Use the product rule to find the derivative of g(x)=x^2 * f(x). The product rule states that (uv)' = u'v + uv', where u=x^2 and v=f(x). So, g'(x) = (x^2)' * f(x) + x^2 * f'(x). Since the derivative of x^2 is 2x, g'(x) = 2x * f(x) + x^2 * f'(x).
02

Compute g'(2)

To find the derivative of g(x) at x=2, substitute x=2 and use the given values f(2)=2 and f'(2)=3 in the g'(x) equation: g'(2) = 2(2) * 2 + (2)^2 * 3 = 4(2) + 4(3) = 8 + 12 = 20.
03

Find equation of tangent line to g(x) at x=2

We know the point of tangency is (2, g(2)), and g(2) = (2)^2 * f(2) = 4 * 2 = 8. The slope of the tangent line is g'(2) = 20. Now, use the point-slope form y - y1 = m(x - x1): y - 8 = 20(x - 2). Thus, the tangent line to y=g(x) at x=2 is y - 8 = 20(x - 2).
04

Find h'(x)

We now need to find the derivative of h(x) = f(x) / (x-3). We apply the Quotient Rule which states that (u/v)' = (u'v - vul) / v^2, where u=f(x) and v=x-3: h'(x) = (f'(x)(x-3) - f(x)) / (x - 3)^2.
05

Compute h'(2)

To find the derivative of h(x) at x=2, substitute x=2, and use the given values f(2)=2 and f'(2)=3 in the h'(x) equation: h'(2) = (3(2-3) - 2) / (2 - 3)^2 = (-1 - 2) / 1 = -3.
06

Find equation of tangent line to h(x) at x=2

The point of tangency is (2, h(2)), and h(2) = f(2) / (2-3) = 2 / (-1) = -2. The slope of the tangent line is h'(2) = -3. Using the point-slope form y - y1 = m(x - x1) again: y -(-2) = -3(x - 2). Thus, the tangent line to y=h(x) at x=2 is y + 2 = -3(x - 2). #Summary# The equation of the tangent line to y=g(x) at x=2 is y - 8 = 20(x - 2), and the equation of the tangent line to y=h(x) at x=2 is y + 2 = -3(x - 2).

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Key Concepts

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

Product Rule
The Product Rule is essential for finding the derivative of a function where two functions are multiplied together. If you have a function of the form \(g(x) = u(x) \cdot v(x)\), the Product Rule tells you that the derivative \(g'(x)\) is given by:
  • \((uv)' = u'v + uv'\)
In this specific problem, \(u(x) = x^2\) and \(v(x) = f(x)\). This means the derivative, \(g'(x)\), can be found as:
  • \(g'(x) = (x^2)' \cdot f(x) + x^2 \cdot f'(x)\)
  • \((x^2)' = 2x\) gives us \(g'(x) = 2x \cdot f(x) + x^2 \cdot f'(x)\)
By learning how to apply the Product Rule, you can handle complex derivatives involving multiplication of functions with ease.
Quotient Rule
The Quotient Rule is used when you need to differentiate a function that is the quotient of two other functions. Suppose you have \(h(x) = \frac{u(x)}{v(x)}\), then the Quotient Rule helps find \(h'(x)\) using:
  • \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\)
In our problem, \(u(x) = f(x)\) and \(v(x) = x - 3\). Applying the rule gives:
  • \(h'(x) = \frac{f'(x) \cdot (x-3) - f(x)}{(x - 3)^2}\)
This allows you to differentiate quotients accurately and find the slopes of lines tangent to such functions.
Tangent Line
A tangent line is a straight line that just "touches" a curve at a given point. It shows the slope or direction of the curve at that specific point. To find the equation of a tangent line, you need:
  • A point of tangency, \((x_1, y_1)\)
  • The slope at that point
Using the point-slope form \(y - y_1 = m(x - x_1)\), you can create the line's equation. For example:
  • For \(y = g(x)\) at \(x=2\), with a slope of \(20\), it is \(y - 8 = 20(x - 2)\).
  • For \(y = h(x)\) at \(x=2\), with a slope of \(-3\), it is \(y + 2 = -3(x - 2)\).
Tangent lines simplify the understanding of complex curves by providing linear approximations at specific points.
Derivative
A derivative represents the rate of change of a function with respect to a variable. In simpler terms, it shows how a function changes as its input changes. Derivatives are crucial for:
  • Understanding how things like speed, area, and volume change
  • Finding slopes of tangent lines to curves
With a given function, finding its derivative involves rules like the Product Rule and Quotient Rule. For instance, with \(g(x) = x^2 \cdot f(x)\), the derivative \(g'(x)\) is derived using the Product Rule, resulting in \(2x \cdot f(x) + x^2 \cdot f'(x)\). Similarly, for \(h(x) = \frac{f(x)}{x-3}\), we use the Quotient Rule to find \(h'(x)\). Understanding derivatives helps in analyzing and predicting the behavior of various mathematical and real-world phenomena.

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Most popular questions from this chapter

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \frac{2 x}{\left(x^{2}+1\right)^{3}}$$

a. Identify the inner function \(g\) and the outer function \(f\) for the composition \(f(g(x))=e^{k x},\) where \(k\) is a real number. b. Use the Chain Rule to show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\).

A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance \(x\) (in inches) of the mass from its equilibrium position after \(t\) seconds is given by the function \(x(t)=10 \sin t-10 \cos t,\) where \(x\) is positive when the mass is above the equilibrium position. a. Graph and interpret this function. b. Find \(\frac{d x}{d t}\) and interpret the meaning of this derivative. c. At what times is the velocity of the mass zero? d. The function given here is a model for the motion of an object on a spring. In what ways is this model unrealistic?

Find the following higher-order derivatives. $$\frac{d^{n}}{d x^{n}}\left(2^{x}\right)$$

In general, the derivative of a product is not the product of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f g\) equals \(f^{\prime} g^{\prime}\)
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