find the equation

[al-horia]

Hi
Can any one help me in solve this question?

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

 

[srmichael]

Hi
Can any one help me in solve this question?

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

Ok, this is the second question you have asked regarding the same concept. What is the first thing you need to do to the function when solving these types of problems? I don't mean to scare you, but this concept is one of the basic concepts you need to understand in calculus. Has your teacher explained how to do this to the class in an understandable manner?

 

[Deleted member 4993]

Hi
Can any one help me in solve this question?

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

Hint:

1) find the derivative of the given function

2) Evaluate the dervative at the given 'x'

and continue....

Please read the post titled "Read before Posting".

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[al-horia]

Hi
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

 

[JeffM]

Hi
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) OK So far.

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 Uh oh This is not correct. See below.
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

The chain rule is your friend. Cherish it.

\(\displaystyle y = \left(- \frac{1}{3}\right)x^3 + \sqrt{8x}.\)

\(\displaystyle Let\ u = - \frac{1}{3}x^3 \implies \dfrac{du}{dx} = - x^2.\)

\(\displaystyle Let\ v = 8x \implies \dfrac{dv}{dx} = 8.\)

\(\displaystyle Let\ w = \sqrt{v} = v^{(1/2)} \implies \dfrac{dw}{dv} = \frac{1}{2}v^{-(1/2)} = \dfrac{1}{2\sqrt{v}}.\)

\(\displaystyle So\ y = u + w \implies \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dw}{dx} = - x^2 + \dfrac{dw}{dv} * \dfrac{dv}{dx} =\)

\(\displaystyle -x^2 + \left(\dfrac{1}{2\sqrt{v}} * 8\right) = -x^2 + \dfrac{4}{\sqrt{8x}}.\)

Now give it a shot.

 

[srmichael]

Hi
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) ===> Correct!

then we need to mind slop ===> Ok, I had a chuckle when I saw this typo
m=f'(x)=f'(2)
f'(x)= -1/3 * 3 x^2 + 1/2 (8x)^-1/2 ===> INCORRECT!! Must use Chain Rule
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

As I stated above, you did not apply the Chain Rule properly in your derivative. Try again and then we'll reconvene.

 

[al-horia]

The chain rule is your friend. Cherish it.

\(\displaystyle y = \left(- \frac{1}{3}\right)x^3 + \sqrt{8x}.\)

\(\displaystyle Let\ u = - \frac{1}{3}x^3 \implies \dfrac{du}{dx} = - x^2.\)

\(\displaystyle Let\ v = 8x \implies \dfrac{dv}{dx} = 8.\)

\(\displaystyle Let\ w = \sqrt{v} = v^{(1/2)} \implies \dfrac{dw}{dv} = \frac{1}{2}v^{-(1/2)} = \dfrac{1}{2\sqrt{v}}.\)

\(\displaystyle So\ y = u + w \implies \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dw}{dx} = - x^2 + \dfrac{dw}{dv} * \dfrac{dv}{dx} =\)

\(\displaystyle -x^2 + \left(\dfrac{1}{2\sqrt{v}} * 8\right) = -x^2 + \dfrac{4}{\sqrt{8x}}.\)

Now give it a shot.

thank you very much

so the slope will be -3
I get the right answer

 

[lookagain]

Hi
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 You must use grouping symbols around the exponent, as in (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

It's the same here with this exponent. Also, the parentheses around the fractional coefficient reads better.
Make it (1/2)(8x)^(-1/2) instead.


f'(x)=-x^2 + 1/2 (8x)^-1/2 Make it (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

Also, yiur work would be more easily read if you add some horizontal spaces
between certain lines.