f(x)=sqrt x find tangent line at point (4,2)?

Question

f(x)= square root (x)
find tangent line at point (4,2)
Please show your work and explain how you got your answer

Answer 1

I'll tell you how to do it, then you can work out the answer.

The tangent of any curve at a given point is the value of the first derivative at that point.

So first you need to differentiate f(x) = sqrt(x) to get f'(x)
[ Hint: sqrt(x) = x^(1/2) ]

Then just evaluate f'(4)

Answer 2

First differentiate the function
f(x) = x^1/2
f'(x) = 1/2 x^-1/2
So when 'x' is 4 substitute '4' into the differentiated eq'n
f'(4) = 1/2 4^-1/2
f'(4) = 1/2 x 1/2 = 1/4

NB 4^-1/2 = 1/4^1/2 = 1/2

So the slope of the tangent is 1/4
Using the straight line eq'n y = mx + c
Then y = 1/4x + c
Substitue in the coordinates for the point (4,2)
2 = 1/4(4) + c
c = 2 - 1 = 1
so tangent line is
y = 1/4x +1 - the answer.

Answer 3

Tangent is equal to f'(x).
f'(x) = 1/[2*square root (x)]
so tangent line at point (4,2) is :
f'(4) = 1/[2*square root (4)] = 0.25

Answer 4

Your function is extra suitable written as (a million/2)e^x the element of tangency is (Ln(3), 3/2) y' = (a million/2)e^x confident, this is the comparable function So the slope of the tangent line at (Ln(3), 3/2) is y = 3x/2 + 3/2(a million - Ln(3))

Answer 5

f(x)=√x --> f'(x)=1/[2·√x]

a=4
b=f(4)=√4 =2
m=f'(4)=1/[2·√4] = 1/4--->

Tangent: y-b=m(x-a)
y-2=1/4·(x-4)
4y-8=x-4
x-4y+4=0 or y=1/4·x +1

Saludos.

Answer 6

f(x) = SQRT (x) = x^(1/2)

f ' (x) = (1/2) x^(-1/2)

m = f ' (4) = (1/2)*(4)^(-1/2) = (1/2)/(4)^(1/2) = (1/2)/2 = 1/4

y - 2 = (1/4)(x - 4)
y - 2 = (1/4)x - 1
y = (1/4)x +1