Question

Help please!

Answer 1

fx may be 2x^2 + 6 as when x=1, fx = 8 and when x=1, f'x = 2x = 2

so 3 times that
=6x^2 +18

can u do the rest?

Answer 2

What do I do next? I'm kinda lost atm

Answer 3

h(x) = sqrt (4+(3(fx)))
f(x) = 2x^2 + 6
3f(x) = 6x^2 +18
4 stays the same, so just put them all together

Answer 4

h(x) = sqrt (6x^2 +22)
h'x = 6x / sqrt (6x^2 +22)
and when x =1
...

Answer 5

so sub in 1 wherever you see an x
it should eventually reduce down to h'x = 3/ sqrt 7

Answer 6

sorry *h'(1) = 3/ sqrt 7

Answer 7

according to webwork that is wrong =/

Answer 8

damn... k what does "webwork"say is right?

Answer 9

it doesn't tell me the answer if I get it wrong but I got it anyways - the answer is 3/(28)^(1/2)

Answer 10

@jim_thompson5910 @Jhannybean could you check this if you get a sec, maybe my maths is off

Answer 11

shouldnt that be 6 / sqrt 28 ... which further reduces to 3/ sqrt 7...?

Answer 12

h(x) = sqrt(4 + 3*f(x))

h ' (x) = 1/(2*sqrt(4 + 3*f(x))) * d/dx(4 + 3*f(x))

h ' (x) = 1/(2*sqrt(4 + 3*f(x))) * (3*f ' (x))

h ' (1) = 1/(2*sqrt(4 + 3*f(1))) * (3*f ' (1))

h ' (1) = 1/(2*sqrt(4 + 3*8)) * (3*2)

h ' (1) = 1/(2*sqrt(28)) * 6

h ' (1) = 6/(2*sqrt(28))

h ' (1) = 3/sqrt(28)

h ' (1) = 3/(2*sqrt(7))

h ' (1) = (3sqrt(7))/(2*7)

h ' (1) = (3sqrt(7))/14

Answer 13

ah ha, didnt use the product rule, my bad
thanks @jim_thompson5910

Answer 14

yw