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**naturewild****Guest**

After spending so much time on these two questions, I just simply gave up. But I couldnt concentrate on other things knowing these questions are unsolved.

Here's the question :

1. f(x) = 2 - x² g(x) = 5 + x²

Find all the common tangent lines equation.

2. Look at the first quadrant in the function 1/x. Let P be any point on the graph. The tangent line at P crosses the x-axis at A and the y-axis at B. O is the origin. Show that the Area of triangle AOB is 2.

Thank you in advance.

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

1. To find gradients of lines, you need to differentiate.

f'(x) = -2x, g'(x) = 2x.

Equating these gives that the only solution is when x = 0. At that point, the gradients of both are 0.

2. The gradient of any point on the curve is -1/x². So, point P has co-ordinates of (x, 1/x) and a gradient of -1/x².

With a line of gradient g and a point of (x,y), the y-intercept is y - (g*x) and the x-intercept is x - (y/g)

So, the y-intercept is 1/x - (-1/x²*x) = 1/x - (-1/x) = 2/x and the x-intercept is x - (1/x ÷ -1/x²) = x - (-x) = 2x.

The x-intercept represents the base length of triangle AOB and the y-intercept represents its height.

Therefore, the triangle's area can be found by multiplying them and dividing by 2.

Area = [2x*(2/x)]/2 = 4/2 = 2.

Why did the vector cross the road?

It wanted to be normal.

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