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Maths
A Level

Find two values of k, such that the line y = kx + 2 is tangent to the curve y = x^2 + 4x + 3

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...

Answered by Andrew N. Maths tutor
33770 Views

Given f(x) = 3 - 5x + x^3, how can I show that f(x) = 0 has a root (x=a) in the interval 1<a<2?

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) + 13<...

Answered by Giorgos P. Maths tutor
6979 Views

(1.) f(x)=x^3+3x^2-2x+15. (a.) find the differential of f(x) (b.) hence find the gradient of f(x) when x=6 (c.) is f(x) increasing or decreasing at this point?

(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.

Answered by Joel E. Maths tutor
3245 Views

The equation x^3 - 3*x + 1 = 0 has three real roots; Show that one of the roots lies between −2 and −1

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) ...

Answered by Paul T. Maths tutor
10511 Views

y=4x^3+6x+3 so find dy/dx and d^2y/dx^2

dy/dx=12x^2+6 d^2y/dx^2=24x

Answered by Swapnil P. Maths tutor
4559 Views

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