Hi! I'm not entirely sure what I'm doing wrong. y=(x+3)^1/2 at x=-3
Now how do I find the equation of the line tangent to that? Critical Points= (-3,0)

Question

Hi! I'm not entirely sure what I'm doing wrong. y=(x+3)^1/2 at x=-3
Now how do I find the equation of the line tangent to that? Critical Points= (-3,0)

mathmale · Answer

$$y=(x+3)^{1/2}$$

anonymous · Answer

@mathmale ahh sorry I had a typo it should be (x+3)^1/3 not ^1/2 sorry!

anonymous · Answer

Derivative is 1/3(x+3)^2/3

mathmale · Answer

Find the derivative of this to obtain an expression in x for the slope of the tangent line.  Evaluate this derivative IF POSSIBLE at -3.  If not possible, what does that tell you?  I don't agree that x=0 is a critical value of this function.

If you mean an EXPONENT of 1/3, please enclose that "1/3" within parentheses for added clarity.

anonymous · Answer

y=x+3^(1/3)

mathmale · Answer

I can't overstress how important it is that you use parentheses in operations such as these.  

Your original function should be typed as y=(x+3)^(1/3) (note the parentheses).
Your derivative should be typed as

(dy/dx) = (1/3)(x+3)^(-2/3).

mathmale · Answer

What is your critical value?  your critical point?

mathmale · Answer

To finish this problem:
1.  find the critical value (there is just one such value).
2.  Evaluate the function at this critical value of x.  This will give you the coordinates of the point of tangency of your tangent line.
3.  Evaluate the derivative, (dy/dx) at this critical value.  If, for any reason, you can't do that, what would that fact tell you?

4.  Put this info together to write the equation of the tangent line.

I need to get off the 'Net; sorry.