find the equation - have error in the final answer !!

[al-horia]

Hi
Can any one help me to get the final answer correct

find the equation of the line that is tangent to the graph of the given function at the point (c, f(c) ) for the specified value of x = c

f(x) = -1/3 x^3 + square root ( 8x ) ; x = 2

in my question
I know how to solve it but i did not get the right answer
f(x) = -1/3 x^3 + square root ( 8x ) ; x = 2

f(x) = -1/3 x^3 + (8x)^1/2
to find y I do this
f(2) = -1/3 * 2^3 + (8*2)^1/2
f(2) = 4/3
points (2,4/3)

then we need to mind slop
m=f'(x)=f'(2)
f'(x)= -1/3 * 3 x^2 + 1/2 (8x)^-1/2
f'(x)=-x^2 + 1/2 (8x)^-1/2
f'(2) = -31/8 = m

the equation
y-y1=m(x-x1)
y-4/3=(-31/8)x + 31/4
y=(-31/8)x +109/12

the right equation is
y=-3x + 22/3

I think my slop is wrong but I don't know how I get the right answer

 

[tkhunny]

f'(x)= -1/3 * 3 x^2 + [1/2 (8x)^-1/2]*8 -- So close. Thou forgettest thy Chain Rule.

 

[Deleted member 4993]

Hi
Can any one help me to get the final answer correct

find the equation of the line that is tangent to the graph of the given function at the point (c, f(c) ) for the specified value of x = c

f(x) = -1/3 x^3 + square root ( 8x ) ; x = 2

in my question
I know how to solve it but i did not get the right answer
f(x) = -1/3 x^3 + square root ( 8x ) ; x = 2

f(x) = -1/3 x^3 + (8x)^1/2
to find y I do this
f(2) = -1/3 * 2^3 + (8*2)^1/2
f(2) = 4/3
points (2,4/3)

then we need to mind slop
m=f'(x)=f'(2)
f'(x)= -1/3 * 3 x^2 + 1/2 * (8)^(1/2) * (x)^-1/2

f'(2) = -4 + 1 = -3


f'(x)=-x^2 + 1/2 (8x)^-1/2
f'(2) = -31/8 = m

the equation
y-y1=m(x-x1)
y-4/3=(-31/8)x + 31/4
y=(-31/8)x +109/12

the right equation is
y=-3x + 22/3

I think my slop is wrong but I don't know how I get the right answer

.

 

[al-horia]

.

thank you very much

I think you had error
f'(2)= -3 not -2

 

[Deleted member 4993]

thank you very much

I think you had error
f'(2)= -3 not -2

You are correct.

I fixed my error in the post.

 

[mmm4444bot]

Hello al-horia:

I merged your two threads together, as the exercise appeared the same in each (except for boldface type).

For future reference, if you would like to change some type to boldface, after you have posted, you may edit your original post. Please do not post duplicate threads. Thank you. :cool: