stumped by calculus Hw, plz. help?

Question

f(x)=Ix2-3xI. Find f'(1)

The equation of the line tangent to y=tan2(3x) at x=p/4 is:

At what point on the curve x3-y2+x2=0 is the tangent line vertical?

If y2-2xy=21, then dy/dx at the point (2, -3) is:

Answer 1

1. f(x) = |x^2 - 3x|
Note that x^2 - 3x > 0 <=> x(x-3) > 0 <=> x < 0 or x > 3.
So we have
f(x) = x^2 - 3x for x ∈ (-∞, 0] ∪ [3, ∞),
and f(x) = -(x^2-3x) = 3x - x^2 for x ∈ [0, 3].
So in the interval (0, 3), f'(x) = 3 - 2x, so f'(1) = 3 - 2 = 1.

2. y = tan^2 (3x)
dy/dx = 2 tan (3x) . sec^2 (3x) . 3 by the chain rule
so at x = π/4 we have dy/dx = 6 tan (3π/4) sec^2 (3π/4)
= 6 (-1) / (-1/√2)^2
= -12.
So the gradient is -12. Also, at x = π/4 we have y = tan^2 (3π/4) = (-1)^2 = 1. So the equation of the line is
y = -12 (x - π/4) + 1
i.e. y = -12x + 3π + 1.

3. x^3 - y^2 + x^2 = 0
Implicit differentiation gives
3x^2 - 2y dy/dx + 2x = 0
The tangent line will be vertical when y is 0 but 3x^2 + 2x is not 0.

Now y^2 = x^3 + x^2 = x^2 (x+1), so y will be 0 when x is 0 or -1. But at x = 0 3x^2 + 2x is also 0, while at x = -1, 3x^2 + 2x = 1. So at (-1, 0) the tangent line will be vertical.

(This analysis doesn't actually demonstrate conclusively that the tangent is not also vertical at x = 0. To do this we have to write y = ± x√(1 + x) and thus show that dy/dx = ±1 at x = 0.)

4. y^2 - 2xy = 21
Implicit differentiation gives
2y dy/dx - 2x dy/dx - 2y = 0
So (y-x) dy/dx = y
So at (2, -3) we have -5 dy/dx = -3, so dy/dx = 3/5.

Answer 2

discover all values t at which the particle is at relax. you're asked while the particle has 0 velocity. At relax = 0 velocity Displacement is given as x(t) = t³ - 12t + 5. spinoff provide you velocity So v(t) = 3t² - 12 Set v(t) = 0 0 = 3t²- 12 0 = t² - 4 0 = (t-2)(t+2) t = ±2 We %. t = 2 in case you placed the displacement equation to 0, you will remedy the time while the displacement is 0,no longer the rate.

Answer 3

at x=1: x^2-3x < 0 ---> Ix^2-3xI = 3x-x^2
f'x = 3-2x
f'(1) = 1