equation of tangent line

[Guest]

Find the equation of the tangent line to the graph of the given fx a the given value of x

f(x)=x^3 sqroot x^3+8 X=1


I really am having a hard time with tangent lines and pointer to help me as well

 

[tkhunny]

Find the derivative.
Evaluate the derivative at the given value. This gives the slope of the line.
Evaluate the original function at the given value. This gives a point on the line.
Use the point-slope form to create the equation.

 

[Guest]

y=1^3 sqrt(1^3+8)=3

y'=x^3(x^3+8)^-1/2(3x^2)+(x^3+8)^1/2 (3x^2)
y'=1(3*1)/sqrt(1+8)+sprt(1+8)(3)=10

y1=3 x1=1 m=10

y=10x-7

but that is not even close to the optional answers I am given and don't know how they get it.

1.19/2x-11/2

2.80/9x-85/9

3. 80/9x+85/9

4. 19/2x-13/2

 

[Gene]

A little off on your f'. Using the prodict rule
f(x)=x^3 sqroot (x^3+8)
f'(x) = x^3d(sqrt(x^3+8))+sqrt(x^3+8)d(x^3)

d(x^3+8)^.5=
(.5)(x^3+8)^-.5*d(x^3+8) =
(.5)(x^3+8)^-(.5)*(3x^2) =
3x^2/(2sqrt(x^3+8))
That makes f' =
3x^5/(2sqrt(x^3+8))+3x^2sqrt(x^3+8)
That will work better.

 

[Guest]

thank you for the that but it still does not give me one of the answers above unless I am doing it wrong.

 

[Gene]

I wouldn't lie to you. What did you get for f'(1).
ps did you make it a fraction, not a decimal?

 

[Guest]

f'=243x/6- 9/3

 

[Gene]

What??????
3x^5/(2sqrt(x^3+8))+3x^2sqrt(x^3+8) =
3*1/(2*sqrt(9))+3*sqrt(9)
I think you've been hitting the sauce :twisted:

 

[galactus]

Find the derivative of the equation to get the slope, as you know.

I got 3x²*sqrt(x³+8)+((3x⁵)/(2sqrt(x³+8)).

Enter x=1 and get m=19/2.

Now, enter x=1 into the original equation to get y.

Now, enter the m value, the x value, and the y value into y=mx+b, solve for b and you have your line equation.

The answer is one of your multiple choices.

 

[tkhunny]

Find the equation of the tangent line to the graph of the given fx a the given value of x
f(x)=x^3 sqroot x^3+8 X=1

Sometimes, it helps to sit back and think about it a little, rather than just ploughing forward. Around x = 1, x > 0. This can be used to our advantage.

f(x) = (x^3)*sqrt(x^3+8) = sqrt(x^9+8*x^6) for x > 1

Now it's only a chain rule problem and not a product rule problem.

f'(x) = (1/[2*sqrt(x^9+8*x^6)])*(9*x^8+48*x^5) which can be simplified quite a bit.

sqrt(x^9+8*x^6) = (x^3)*sqrt(x^3+8) for x > 1 and this is where we started, anyway.

(9*x^8+48*x^5) = 3*(x^5)*(3*x^3+16)

Putting it all back together gives

f'(x) = (1/[2*(x^3)*sqrt(x^3+8)])*(3*(x^5)*(3*x^3+16)) = (3/2)*(x^2)*[(3*x^3+16)/(sqrt(x^3+8))]

It's not a whole lot cleaner, if at all, but it might be sufficiently different to make more sense in some situations.