1. ## Decreasing Functions

Hello!
This is a question about the function f(x) = 1 + e–2x.

First I have to find f¢(x). I have found this to be: -2e^-2x

Then I have to:
Let P be the point on the graph of f where x = –1/2.
(b) Find an expression in terms of e for

(i) the y-coordinate of P;

(ii) the gradient of the tangent to the curve at P.
and finally I have to

(c) Find the equation of the tangent to the curve at P, giving the answer in the form y = ax + b.

Thanks,
Amine

Hello!
This is a question about the function f(x) = 1 + e–2x.

First I have to find f¢(x). I have found this to be: -2e^-2x

Then I have to:
Let P be the point on the graph of f where x = –1/2.
(b) Find an expression in terms of e for

(i) the y-coordinate of P;

(ii) the gradient of the tangent to the curve at P.
and finally I have to

(c) Find the equation of the tangent to the curve at P, giving the answer in the form y = ax + b.

Thanks,
Amine
For part b, just substitute $\displaystyle x=-\frac{1}{2}$, to obtain the function value at the given point. $\displaystyle f(\frac{1}{2})=1+e^{-1}$

This is also the y-coordinate of P. So the point P is $\displaystyle (\frac{1}{2},1+e^{-1})$. This takes care of (i).

The gradient of the curve at P is just the function value of the derivative at P. So $\displaystyle f'(-\frac{1}{2})$. I'll let you evaluate that to finish part (ii).

For part (c) use the formula of a line:

$\displaystyle y-f(a) = f'(a)(x-a)$

Part (b) gave you all of the information you need to fill in this formula. Does this make sense?

Hello!
This is a question about the function f(x) = 1 + e–2x.

First I have to find f¢(x). I have found this to be: -2e^-2x

Then I have to:
Let P be the point on the graph of f where x = –1/2.
(b) Find an expression in terms of e for

(i) the y-coordinate of P;

(ii) the gradient of the tangent to the curve at P.
and finally I have to

(c) Find the equation of the tangent to the curve at P, giving the answer in the form y = ax + b.

Perhaps u have missed the power ^ in your question,

For y cordinate of p,simply substitute x=-1/2 in your origional func;you will get corresponding value of y

For gradient(slope) of tangent; you derivate the func like you have done and put x=-1/2;
value of y=-2e will be your slope;

For equation of tangent,just ut x1,y1 of point P(from part a) and m=slope of tangent (-2e)
in the standard equation of line:

(y-y1)=m(x-x1)