Find equation for the tangent line using the limit definition

[alexmay]

I need assistance with solving this problem and understanding how to work my way through it.

y=x^3/2 (4,128)

 

[lookagain]

y=x^3/2 (4,128)

Whether this is supposed to be y = (x^3)/2 or y = x^(3/2), the point (4, 128) does
not satisfy either.

 

[topsquark]

@alexmay:

Please use a different thread title for individual threads. It can get a bit confusing.

-Dan

 

[alexmay]

I need assistance with solving this problem and understanding how to work my way through it.

y=x^3/2 (8,256)

 

[Romsek]

do you understand that the slope of the tangent line to [MATH]y(x)[/MATH] at some point \(\displaystyle x_0\) is \(\displaystyle \dfrac{dy}{dx}(x_0)\) ?

 

[Steven G]

From what Romsek stated and the point (8,256) can you figure out the equation of the line you want?

 

[Steven G]

Equation should be y = (x^4)/2

 

[Deleted member 4993]

I need assistance with solving this problem and understanding how to work my way through it.

y=x^3/2 (4,128)

Is your problem:

Find equation for the tangent line of:

y = (x⁴)/2

at the point (4 , 128)

using the limit definition

 

[Deleted member 4993]

Please define "tangent line to a curve" using limit definition.

 

[HallsofIvy]

So your problem is to find the tangent line to \(\displaystyle y= \frac{x^4}{2}\), at (4, 128)?

The tangent line is of the form y= m(x- 2)+ b where m is the slope and b is the value of y at x= 4. I presume that "by limits" means that you want to find the slope by using the "limit definition" of the derivative, \(\displaystyle \lim_{h\to 0} \frac{f(4+ h)- f(4)}{h}\)

Taking \(\displaystyle f(x)= \frac{x^4}{2}\), the "difference quotient is \(\displaystyle \frac{(4+h)^4/2- 128}{h}= \frac{128h+96h^2+ 16h^3+ h^4}{h}= 128+ 96h+ 16h^2+ h^3\).

What is the limit of that as h goes to 0?