+118723 Is this half of a question? I'm confused.
Mi tangent at point X = 4 is 1/4 What does this mean?????
ALSO
Do you mean
$$\frac{ax}{\sqrt{x}}-x$$
f(X)= $${\frac{{\mathtt{ax}}}{\left({\sqrt{{\mathtt{x}}}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}}$$
you have the tangent at the point x=4 is 1/4
+118723 ok now it makes a lot more sense. although you missed out the words "the gradient of"
f(X)= $${\frac{{\mathtt{ax}}}{\left({\sqrt{{\mathtt{x}}}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}}$$
you have the gradient of the tangent at the point x=4 is 1/4
The gradient of the tangent of any function is given by the first derivative.
So what you are really being told is that f '(4)=1/4
You have to find f ' (x) (using the product or the quotient rule)
then solve for a using f '(4)=1/4
Can you work out the derivative yourself ? Or can you at least give it a go?
