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Chapter 16: Problem 14

Find the equation of the tangent to \(y=\frac{1}{x^{2}}\) where \(x=-1 .\)

Short Answer

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

Step by step solution

01

Find the Point on the Curve

We are given the \(x\) value as \(x= -1\). We now find the \(y\) value by plugging this into the equation of the curve. This will give us the point \((x,y)\) where the tangent line touches the curve: $$ y = \frac{1}{x^2} = \frac{1}{(-1)^2} = 1 $$ So, the tangent touches the curve at the point \((-1, 1)\).
02

Differentiate the Equation of the Curve

Differentiate \(y = \frac{1}{x^2}\) with respect to \(x\) to obtain the slope function of the curve: $$ y' = \frac{d}{dx} \left( \frac{1}{x^2} \right) = -\frac{2}{x^3} $$
03

Find the Slope of the Tangent at the Given Point

Substitute the given \(x\) value (\(x=-1\)) into the derivative to find the slope of the tangent at the point: $$ m = y'(-1) = -\frac{2}{(-1)^3} = 2 $$ The slope of the tangent at the point \((-1,1)\) is \(2\).
04

Write the Equation of the Tangent Line

Now we can use the point-slope form of a linear equation (\(y-y_1=m(x-x_1)\)) to write the tangent equation using the point \((-1,1)\) and slope \(2\): $$ y-1 = 2(x-(-1)) $$ Simplify the equation to obtain the final tangent equation: $$ y = 2x+3 $$

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Most popular questions from this chapter

Find \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\) given \(y=\mathrm{e}^{\left(x^{2}\right)}\).

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