Question

find the values of ( x ) where the tangent line to the graph of ( f(x)=\frac{1}{x} ) is parallel to the line ( y=-3 x + 4 ). enter the exact values of the answer(s) (not decimal approximations).( x = )

Explanation:

Step1: Find the derivative of \( f(x) \)

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), for \( f(x)=\frac{1}{x}=x^{-1} \), then \( f^\prime(x)=-1\times x^{-2}=-\frac{1}{x^{2}} \).

Step2: Determine the slope of the given line

The line \( y = - 3x+4 \) is in the form \( y = mx + b \) (slope - intercept form), where \( m=-3 \).

Step3: Set the derivative equal to the slope of the line

Since the tangent line is parallel to \( y=-3x + 4 \), their slopes are equal. So we set \( f^\prime(x)=-3 \), i.e., \( -\frac{1}{x^{2}}=-3 \).
Multiply both sides by \( -x^{2} \): \( 1 = 3x^{2} \).
Then \( x^{2}=\frac{1}{3} \).
Take the square root of both sides: \( x=\pm\frac{1}{\sqrt{3}}=\pm\frac{\sqrt{3}}{3} \).

Answer:

\(x = \frac{\sqrt{3}}{3},x=-\frac{\sqrt{3}}{3}\)