Question

estimate the slope of the tangent to the graph of the given function at the point with the given x-coordinate. (Round your answer to three decimal places.)
5*x^1/2 ;x=16

Answer 1

the derivative of 5x^1/2 is 5 * 1/2 x^(1/2 - 1)
= 5/2 x^-1/2
= 5/(2x^1/2)
= 5/2sqrt(x)
now plug x = 16 to equation above, therefore the slope becomes
m = 5/2sqrt(16) = 5/2*4 = 5/8

Answer 2

would the equation of the tangent line be 5/8x+16 ?

Answer 3

we need a point to form the tangen line equation, take x=16 and subtitute to
y = 5x^1/2, so we get
y = 5(16)^1/2 = 5*4 = 20
therefore the tangent line equation of curve y = 5x^1/2 at point (16,20) with the slope m = 5/8 is

use the formula :
y - y1 = m(x-x1)

so, y - 20 = 5/8 (x - 16) or
y - 20 = 5x/8 - 10
y = 5x/8 - 10 + 20
y = 5x/8 + 10

Answer 4

oh ok I understand now. Thank you so much!