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Find the equation of the tangent line to the graph of the given function at the point with the indicated \(x\) -coordinate. In each case, sketch the curve together with the appropriate tangent line. $$ f(x)=\frac{1}{x^{2}} ; x=1 $$

Short Answer

Expert verified
The derivative of the function \(f(x)=\frac{1}{x^{2}}\) is \(f'(x) = -2/x^3\). At \(x = 1\), the slope of the tangent line is \(f'(1) = -2\), and the tangent point is (1, 1). The equation of the tangent line is \(y = -2x + 3\). To sketch the curve and the tangent line, note that \(f(x) = 1/x^{2}\) is a hyperbola with vertical and horizontal asymptotes at \(x = 0\) and \(y = 0\) respectively, and the tangent line \(y = -2x + 3\) is a straight line with slope -2 and y-intercept 3. They intersect at the point (1, 1).

Step by step solution

01

Compute the derivative of the function

The derivative of the function \(f(x)=\frac{1}{x^{2}} = x^{-2}\) can be computed using the power rule of differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\). So, the derivative \(f'(x)\) is \(-2x^{-3} = -2/x^3\).
02

Find the slope of the tangent line and the y-coordinate of the tangent point

Substitute \(x = 1\) into \(f'(x) = -2/x^3\) to find the slope of the tangent line, and into \(f(x) = 1/x^2\) to find the y-coordinate of the tangent point. How to do it: - For the slope, \(f'(1) = -2/1^3 = -2\). - The y-coordinate is \(f(1) = 1/1^2 = 1\). So, the tangent point is (1, 1).
03

Find the equation of the tangent line

The slope-point form of a line is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and m is the slope. Substitute \(m = -2\), \(x_1 = 1\), and \(y_1 = 1\) into the equation to get the equation of the tangent line: \(y - 1 = -2(x - 1)\), which simplifies to \(y = -2x + 3\).
04

Sketch the function and the tangent line

To sketch \(f(x) = 1/x^{2}\), note that it’s a typical hyperbola that opens upward with a vertical asymptote at \(x = 0\) and a horizontal asymptote at \(y = 0\). To sketch the tangent line \(y = -2x + 3\), note that it’s a straight line with slope -2 and y-intercept 3. The tangent line intersects the curve at the point (1, 1). Sketching this accurately would provide a visual representation of the function and its tangent line at \(x = 1\). It's recommended to use graphing tools for accurate sketches.

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Key Concepts

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

Derivative of a Function
When we talk about the derivative of a function, we're referring to the rate at which the function's value changes at a certain point. It gives us the slope of the tangent line to the graph of the function at any point. Imagine you're tracking the position of a car over time—the derivative tells you the car's speed (rate of change) at any instant. In calculus, this concept is foundational, as it is used to understand how functions behave and how they can be optimized.

To find the derivative, we apply rules from calculus, such as the power rule, and techniques for more complex functions involve additional rules and methods. Derivatives are not just abstract; they have practical applications in physics, economics, biology, and nearly every field of science and engineering. They help in understanding and predicting the behavior of various phenomena by modeling how variables change in relation to one another.
Slope-Point Form of a Line
The slope-point form of a line is a way to write the equation of a line when you know its slope and one point on the line. Mathematically, it's expressed as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the known point and \(m\) is the slope. This form is extremely useful because it allows us to quickly write an equation as soon as we know how steep the line is (the slope) and where it is positioned (the point).

In practical terms, think of it as a starting point and direction for drawing a line on a graph. If you're given a point through which the line passes and how steeply it inclines or declines (slope), you can lay out its entire course. Moreover, it's a go-to method when dealing with tangent lines, as these lines merely 'touch' the function at a single, precise point.
Power Rule of Differentiation
The power rule is one of the most straightforward and commonly used rules for finding derivatives. It states that if you have a function where the variable, say \(x\), is raised to a power, like \(x^n\), the derivative of that function is \(nx^{n-1}\). Essentially, you bring down the exponent as a coefficient (multiply), then subtract one from the exponent.

This rule simplifies the process of differentiation, especially for polynomial functions, which are sums of terms like \(x^n\). Whether the exponent is positive or negative, the power rule applies, thereby it's a go-to tool for finding slopes of tangent lines to all sorts of curves, not only simple straight lines but also the peaks and valleys of more complex shapes.

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Most popular questions from this chapter

If f and g are functions of time, and at time t = 3, f equals 5 and is rising at a rate of 2 units per second, and g equals 4 and is rising at a rate of 5 units per second, then f/g equals ____ and is changing at a rate of ____ units per second.

Compute the indicated derivative using the chain rule. HINT [See Quick Examples on page 828.] \(x=1-t / 2, y=4 t-1 ; \frac{d y}{d x}\)

Find the equation of the line tangent to the graph of the given function at the point with the indicated \(x\) -coordinate. \(f(x)=\left(x^{0.5}+1\right)\left(x^{2}+x\right) ; x=1\)

The percentage y (of total personal consumption) an individual spends on food is approximately y = 35x?0.25 percentage points (6.5 ? x ? 17.5) where x is the percentage the individual spends on education.28 An individual finds that she is spending x = 7 + 0.2t percent of her personal consumption on education, where t is time in months since January 1. Use direct substitution to express the percentage y as a function of time t (do not simplify the expression) and then use the chain rule to estimate how fast the percentage she spends on food is changing on November 1. Be sure to specify the units.

The following graph shows the approximate value of home prices and existing home sales in 2006–2010 as a percentage change from 2003, together with quadratic approximations The quadratic approximations are given by $$ \text { Home Prices: } \quad P(t)=t^{2}-10 t+41 \quad(0 \leq t \leq 4) $$ Existing Home Sales: \(\quad S(t)=1.5 t^{2}-11 t \quad(0 \leq t \leq 4)\) where \(t\) is time in years since the start of 2006 . Use the chain rule to estimate \(\left.\frac{d S}{d P}\right|_{t=2} .\) What does the answer tell you about home sales and prices? HINT [See Quick Example 2 on page 828.]

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