Slope of tangent

[marko02]

I am trying to figure out this equation but I am stuck. The question is as follows: Consider the functon f(x) = x sqroot(20-x) , find the slope of the tangent when x=11

I've worked the f(x+h) -f(x) to (11+h)sqroot(9-h) - 33 x sqroot(9-h) +33
aaaaaaaaaaaaaaaa haaaaaaaaaaaaahaaaaaaaaaaaaaaaasqroot(9-h) +33

Everytime I multiply in I keep getting stuck how to simplify and keep getting the wrong answer. Any help would be appreciated.

Thanks

 

[srmichael]

I am trying to figure out this equation but I am stuck. The question is as follows: Consider the functon f(x) = x sqroot(20-x) , find the slope of the tangent when x=11

I've worked the f(x+h) -f(x) to (11+h)sqroot(9-h) - 33 x sqroot(9-h) +33
aaaaaaaaaaaaaaaa haaaaaaaaaaaaahaaaaaaaaaaaaaaaasqroot(9-h) +33

Everytime I multiply in I keep getting stuck how to simplify and keep getting the wrong answer. Any help would be appreciated.

Thanks

That's because the conjugate is \(\displaystyle \displaystyle (11+h)\sqrt{(9-h)}+33\)

 

[marko02]

That's because the conjugate is \(\displaystyle \displaystyle (11+h)\sqrt{(9-h)}+33\)

Thanks, so would I be correct in the next step being

\(\displaystyle \displaystyle (11+h)^2\((9-h)-1089/(h(11+h)\sqrt{(9-h)}+33)\)

And if so I am still lost as how to simplify this

 

[srmichael]

Thanks, so would I be correct in the next step being

\(\displaystyle \displaystyle (11+h)^2\((9-h)-1089/(h(11+h)\sqrt{(9-h)}+33)\)

And if so I am still lost as how to simplify this

Yes.

\(\displaystyle \displaystyle \frac{(11+h)^2(9-h)-1089}{h[(11+h)\sqrt{(9-h)}+33]}\)

\(\displaystyle \displaystyle \frac{(121+22h+h^2)(9-h)-1089}{h[(11+h)\sqrt{(9-h)}+33]}\)

\(\displaystyle \displaystyle \frac{(1089+198h+9h^2-121h-22h^2-h^3)-1089}{h[(11+h)\sqrt{(9-h)}+33]}\)

\(\displaystyle \displaystyle \frac{77h-13h^2-h^3}{h[(11+h)\sqrt{(9-h)}+33]}\)

\(\displaystyle \displaystyle \frac{h(77-13h-h^2)}{h[(11+h)\sqrt{(9-h)}+33]}\)

\(\displaystyle \displaystyle \frac{77-13h-h^2}{(11+h)\sqrt{(9-h)}+33}\)

Now, \(\displaystyle h\rightarrow 0\).

Have at it!

 

[marko02]

Perfect thanks! I kept trying to cancel out numbers, kept messing me up.