Find the equation of the line that is tangent to the graph of the function [tex]f(x)=\ln (9 x+5)[/tex] at the point [tex](0, \ln 5)[/tex]. Put your answer in the form [tex]y=m x+b[/tex].
Answer (1)
Find the derivative of the function f ( x ) = ( 9 x + 5 ) , which is f ′ ( x ) = (( 9 x + 5 )) .
Evaluate the derivative at x = 0 to find the slope: m = f ′ ( 0 ) = (( 5 )) .
Use the point-slope form to find the equation of the tangent line: y − 5 = (( 5 )) ( x − 0 ) .
Rewrite the equation in slope-intercept form: y = (( 5 )) x + 5 .
Explanation
Problem Setup We are given the function f ( x ) = ( 9 x + 5 ) and the point ( 0 , 5 ) . We need to find the equation of the tangent line to the graph of f ( x ) at this point. The equation of a tangent line is given by y = m x + b , where m is the slope and b is the y-intercept.
Finding the Derivative First, we need to find the derivative of f ( x ) to determine the slope of the tangent line. Using the chain rule, we have f ′ ( x ) = (( 9 x + 5 ) ) ′ = (( 9 x + 5 )) (( 9 x + 5 ) ) ′ = (( 9 x + 5 )) ( 9 ) = (( 9 x + 5 )) .
Calculating the Slope Next, we evaluate the derivative at x = 0 to find the slope m at the point ( 0 , 5 ) :
m = f ′ ( 0 ) = (( 9 ( 0 ) + 5 )) = (( 5 )) = (( 5 )) . So, m = (( 5 )) .
Finding the Tangent Line Equation Now we use the point-slope form of a line, y − y 1 = m ( x − x 1 ) , with the point ( 0 , 5 ) and the slope m = (( 5 )) :
y − 5 = (( 5 )) ( x − 0 ) y = (( 5 )) x + 5.
Final Equation Thus, the equation of the tangent line is y = (( 5 )) x + 5 .
Examples
Imagine you are designing a curved ramp for a skateboard park. To ensure a smooth transition at a specific point on the ramp, you need to find the tangent line to the curve at that point. This tangent line helps you determine the angle and direction of the ramp at that location, ensuring a safe and enjoyable ride for skateboarders. The principles of finding tangent lines are crucial in engineering design to create smooth and functional structures.
