if f(x) = 1/x, then since
1/x = x^-1, f' = (-1)x^-2 = -1/x^2
at x=1, that is -1/a^2
huh? ((2/a)(2a))/2 = (1/2)(2/a)(2a)
dividing by 2 is the same as multiplying by 1/2, right?
Let (a, b) be any point on the graph of Prove that the area of the
triangle formed by the tangent through (a, b) and the coordinate axes is 2.
This is the solution from the solution manual
The coordinates of the point are (a,1/a). The slope
of the tangent is -1/a^2. The equation of the tangent
is y-1/a=-1/a^2(x-a) or y=(-1/a^2)x + 2/a. The
intercepts are (0,2/a) and (-2a,0). The tangent line
and the axes form a right triangle with legs of length 2/a and 2a. The area of the triangle is (1/2)(2/a)(2a)=2.
I understand everything about this except I don't know how the slope of the tangent is -1/a^2 was calculated. Also, I'm confused about the area of the triangle equation. Shouldn't it be Area=(height x base)/2 so ((2/a)(2a))/2 which is different from (1/2)(2/a)(2a)=2.
3 answers
6. The following are the x-intercepts, EXCEPT
a. -5 b. c. d. 2
a. -5 b. c. d. 2
7. Which of the following is the y-intercept?
a. -120 b. -60 c. 60 d. 120
a. -120 b. -60 c. 60 d. 120