Determine the equation of the tangent line

Blablablabla
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Homework Statement



\frac{3x+6}{2-x}

at x=3


Homework Equations



y - y_{o} = m(x-x_{o})

The Attempt at a Solution



f(3) = -\frac{15}{4}

m = \frac{3}{0} DNE



I have to write the equation in the form of the point-slope formula.

I can get x_{o} and y_{o}, but I am having trouble finding m.

Thanks for any help.
 
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You didn't give an equation. I suppose it is$$
y=\frac{3x+6}{2-x}$$You need to calculate its derivative at ##3## to get the slope. You might also check your ##f(3)##.
 
Sorry, yes that is the equation. Can you help me find the derivative? I'm a bit confused because my textbook says that the derivative is

\frac{f(x+h)-f(x)}{h}

but in class we learned that as the difference quotient, and that the derivative is when you do this:

y = x^{n}
y' = nx^{n-1}

Thanks for the fast reply
 
Blablablabla said:
Sorry, yes that is the equation. Can you help me find the derivative? I'm a bit confused because my textbook says that the derivative is

\frac{f(x+h)-f(x)}{h}

That is not the derivative of f(x). You have to take the limit as ##h \to 0## to get the derivative.
but in class we learned that as the difference quotient, and that the derivative is when you do this:

y = x^{n}
y' = nx^{n-1}

Thanks for the fast reply

That gives the rule for differentiating powers, which is derived from the difference quotient by letting ##h\rightarrow 0##. For more complicated derivatives like your quotient, you would use the quotient rule and the power rule. Haven't you had the quotient rule?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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