Refreshers and/or solutions

[lilrockfreekshow]

Ok here's the first problem

Let f be the function defined by f(x)=(1+tan x)^(3/2)

a) Write an equation for the line tangent to the graph of f at the point where x=0
b)Using the equation found in part (a), approximate f(0.02)
c)Let f^-1 denote the inverse function of f. Write and expression that gives f^-1(x) for all x in the domain of f^-1

Someone tell me what to use exactly to do these. I don't need the actual solution, just how to do it.

 

[pka]

Here are some hints,
\(\displaystyle \L
f'(x) = \frac{3}{2}\left( {1 + \tan (x)} \right)^{\frac{1}{2}} \left( {\sec ^2 (x)} \right)\)


\(\displaystyle \L
\begin{array}{c}
x = \left( {1 + \tan (y)} \right)^{\frac{3}{2}} \\
x^{\frac{2}{3}} = 1 + \tan (y) \\
y = \arctan \left( {x^{\frac{2}{3}} - 1} \right) \\
\end{array}\)

 

[lilrockfreekshow]

Ok well since I can't figure this one out and I have a whole packet of ten problems, all different. I give up, I'll let my teacher explain them. I'll go after school for help or something.

Thanks for your hints though.

 

[galactus]

Don't give up that easy!. pka gave you the derivative(f'(x)) of the function. That's

the slope. You know, the m in y=mx+b.

They want the slope at x=0. Enter x=0 into the derivative and see what you get.

Enter x=0 into the original function to get y, then set up your line equation and

solve for b. You know x=0 already.

 

[lilrockfreekshow]

Now I remember... wow... 2 weeks of vacation and my brain goes... pll...

Thank you so much, now that makes things clearer on the other problems too!