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Firstly remember that each part of the equation can be differentiated separately. Let's label each part:

Part A: 2(x^2)y

Part B: 2x

Part C: 4y

Part D: -cos(piy)

Part E: ...

There will be intersection when x^2 + 4x + 3 = kx + 2. Our goal is to find the values of k which would only give one solution to this quadratic equation, which would make the lines 'tangent' to each other...

In plain english, we need to show that there is a value of x, which we call "a", in the interval 1 < a < 2 where f(a)=0. To prove this we start by letting x = 1: f(1) = 3 - 5(1) + 1^3<...

(1.) (a.) f’(x)=3x^2+6x-2

(b.) x=6 gradient=142

(c.) since f’(x)>0 at x=6, the function is increasing.

In order to prove that one real root of an equation is situated in a certain interval, we calculate the value of the function at the ends of the given interval. In the given case, f(-2) = (-2)^3 - 3*(-2) ...