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Chapter 3: Problem 32

Derivatives and tangent lines a. For the following functions and values of \(a,\) find \(f^{\prime}(a)\) b. Determine an equation of the line tangent to the graph of \(f\) at the point \((a, f(a))\) for the given value of \(a\) $$f(x)=\frac{1}{x^{2}} ; a=1$$

Short Answer

Expert verified
Answer: The equation of the tangent line to the curve \(f(x) = \frac{1}{x^2}\) at the point \((1, 1)\) is \(y = -2x + 3\).

Step by step solution

01

Find the first derivative of the function.

To find the first derivative of the function \(f(x) = \frac{1}{x^2} = x^{-2}\), we will apply the power rule (\(\frac{d}{dx}(x^n) = nx^{n-1}\)). $$f^{\prime}(x) = -2x^{-2 - 1} = -2x^{-3} = \frac{-2}{x^3}$$
02

Evaluate the first derivative at the given value of \(a\).

Now that we have the first derivative, we can evaluate it at the given value of \(a=1\) to find the slope of the tangent line at that point. $$f^{\prime}(1) = \frac{-2}{1^3} = -2$$ So the slope of the tangent line to the curve at \(a=1\) is \(-2\).
03

Find the function value at the given point.

Next, we need to find the value of \(f(a)\), i.e., \(f(1)\). $$f(1) = \frac{1}{1^2} = 1$$ Thus, the point on the curve where the tangent line touches is \((1, 1)\).
04

Determine the equation of the tangent line using point-slope form.

Now we have the point \((1,1)\) and the slope \(-2\). Using the point-slope form of a straight line, which is given by \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is the point and \(m\) is the slope, we can find the equation of the tangent line. After substituting values, we get: $$y - 1 = -2(x - 1)$$ Simplify the equation to get the final answer: $$y = -2x + 3$$ Therefore, the equation of the tangent line to the curve \(f(x) = \frac{1}{x^2}\) at the point \((1, 1)\) is \(y = -2x + 3\).

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Most popular questions from this chapter

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. \(x y=a ; x^{2}-y^{2}=b,\) where \(a\) and \(b\) are constants

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