Find k such that y = 4x + 4 is tangent to f(x) = k*sqrt(x)

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Find k such that y = 4x + 4 is tangent to f(x) = k*sqrt(x). Very nice Calculus problem where you have to think.

[@nicolekuperman917]

I just found your videos, and I cannot tell you how helpful these are. Clearly explained with simple steps.

[@thiagootto484]

Bless your soul!

[@TheMathSorcerer]

lol thanks:)

[@Ballergotswag]

Thank you!

[@TheMathSorcerer]

You are welcome!

[@PsychoticAnchovies]

you're a life-saver, man! thanks a bunch!

[@TheMathSorcerer]

hey np:)

[@nicolespinelli4833]

This tutorial helped so much, thank you!

[@rnelson_27]

this was so incredibly helpful THANK YOU!!

[@JamesBrodski]

Thank you so much! Very helpful.

[@TheMathSorcerer]

You are welcome!!

[@TheDragonKing444]

Thank you homie

[@TheMathSorcerer]

Np 😄

[@epb111]

is k a constant? Why don't we have to use the product rule for derivatives?

[@TheMathSorcerer]

[@amp1207]

Thanks 🙏🏾

[@EthanLeavitt]

Thank you

[@kleodavelajarca7479]

THANK YOU SO MUCH :)

[@TheMathSorcerer]

glad it helped!!!

[@alexthompson9409]

What if you have k-x^2 tho? Finding it's derivative would make k disappear because it is a constant so how can you solve for k?

[@runeyugi]

+Alex Thompson (Eaglesfan1825) I'm trying to figure out this as well. If you solved K - x^2 for k I would love to know how you did that.

[@donovanvaughan6053]

I too would like this answered

[@diegomiranda3243]

Larson Calculus Eighth Edition? 😂 I'm trying to solve this one as well.

[@fermidirac8904]

you take the derivative to find the point of tangency and when you have that you plug in the coordinate points back into the original function, not the derivative, and then you can solve for k

[@arushjain1299]

tysm

[@user-up4wj9vi3w]

thanks

[@GruhChannel]

thanks cuh

[@sonyx7138]

thanks man

[@mehrdadheydari5025]

thank u a lot

[@TheMathSorcerer]

np:)

[@negarfarahbakhsh4890]

Thank you!

[@TheMathSorcerer]

You are welcome!