Sole this equation?

Question

Find a line tangent to the curve y = X^(1/2)

at x=4

Answer 1

f(x) = x^(1/2)
f `(x) = (1/2) x ^(-1/2)
f `(x) = 1 / (2 √ x )
f `(4) = 1 / 4 = m
Tangent touches (4 , 2)
y - 2 = (1/4) (x - 4)
y - 2 = (1/4) x - 1
y = (1/4) x + 1

Answer 2

We find the gradient of the line dy/dx to get 1/2x^1/2. Sub 4 into it to get 1/2(4)^1/2 = 1/4.
y = 1/4x.
Sub the x =0 to get the y intercept.
y = 1/4(0)
y =0
So the line tangent is y= 1/4x

Answer 3

this is differiantiation. when u differentiate an eqn (make it dy/dx), you find the eqn of the gradient.

so with y = x^(1/2)
differentiation is bring down the power, power -1 in that order. so by bringing down the power, the coefficient of x becomes 1 x 1/2, so is now 1/2, and then power -1, so the power is now -1/2
dy/dx = 1/2 x^(-1/2)

and with this eqn u can find the gradient at any point u want on the curve. but since they want x=4, sub it in
1/2(4^-1/2)
a negative power denotes its 1/(whatever) and 1/2 denotes squareroot.
so 1/2 (4^-1/2) = 1/2 ( 1/ sqrt4)
= 1/2 (1/2)
=1/4

Answer 4

y = x^1/2, so

dy/dx = 1/2 x^-1/2

=1/2rtx

When x = 4,

dy/dx = 1/2rt4

= 1/2*2

= 1/4.

So, using y = mx + c, (m = 1/4 and x = 4, y = 2) then

y = x/4 + c, so

2 = 4/4 + c, so

2 = 1 + c, so

c = 1, and

y = x/4 +1.

Hope this helps, Twiggy.

Answer 5

y=x^(1/2)
dy/dx=1/2x^(1/2).
when x=4, y=2
dy/dx=1/2*2
=1/4.
the equation of the tangent at (4,2) is:
y-2=1/4(x-4).ANS

Answer 6

y = X^(1/2)

dy/dx=1/2x^-1/2=1/2(4)^-1/2=1/2*1/(4)^1/2
=1/2*1/2=1/4

Answer 7

This is the most basic type of question that can be asked in calculus. Unless you learn to solve problems like these by yourself (preferably in your sleep), you are not going to make a passing grade. Better turn off the computer, open your textbook and start studying.

Answer 8

Z=-3.5