Question

find an equation of the line that is tangent to the graph of the function f(x)=7/(√x) and parallel to the line 7x+2y-6=0

Answer 1

first, can you determine the slope of line 7x+2y-6=0 ?

Answer 2

@raden 7/2?

Answer 3

the slope of line ax + by + c = 0 is
m = -a/b
(with m is slope)
see the equation 7x+2y-6=0, it means a = 7, b = 2
therefore m = -a/b = -7/2

Answer 4

ok

Answer 5

then we knowed that 2 lines are parallel if the slope is same. so use m = -7/2 too

Answer 6

now find derivative of f(x) = 7/(√x) = 7/x^1/2 = 7x^(-1/2)
f ' (x) = -1/2 * 7 * x^-3/2 = -7/2 x^(-3/2) = -7/(2x^3/2)

because the slope (m) = -7/2, then -7/(2x^3/2) = -7/2
solve for x = ........ ?

Answer 7

I got x=1 but thats not an answer choice.

ANSWER CHOICES:
3x+y-6=0
7x+2y-14=0
7x+2y-21=0
3x+2y+3=0
3x+y+21=0

Answer 8

@RadEn

Answer 9

yeah, it just point slope coordinat, not tangent equation :)
now subtitute x = 1 into f(x)=7/(√x) to get y

Answer 10

y = 7/(√1) = 7,\ so it is (1, 7) as the point slope

Answer 11

ok

Answer 12

finally, use the formula of tangent line :
y - y1 = m (x - x1)
y - 7 = -7/2 (x - 1)

times 2 on both sides, we get
2(y - 7) = -7(x - 1)
2y - 14 = -7x + 7

add 7x on both sides :
7x + 2y - 14 = 7x -7x + 7
7x + 2y - 14 = 7

subtracted by 7, on both sides :
7x + 2y - 14 - 7 = 7 - 7
7x + 2y - 21 = 0

Answer 13

thanks!

Answer 14

you're welcome ;)