Find the equation of a curve that passes through the point (1,0), has the tangent 6x + y = 6 at that point, and has a second derivative y" = 12/(x-3) .
Added by Danielle F.
Step 1
Step 1: Find the first derivative y'** Given: y" = 12/(x-3) Integrating y" with respect to x, we get: y' = ∫(12/(x-3)) dx y' = 12 ln|x-3| + C ** Show more…
Show all steps
Close
Your feedback will help us improve your experience
Adi S and 57 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the equation of the tangent line to the curve `y=6 sqrt(x) + x` when `x=9`. Use point-slope form.
Zhumagali S.
Find an equation of the straight line tangent to the given curve at the point indicated. $$y=6-x-x^{2} \text { at } x=-2$$
Differentiation
Tangent Lines and Their Slopes
tangent line equation
Atul K.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD