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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) + 1^3<...

Answered by Giorgos P. • Maths tutor

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(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

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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

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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

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The straight line L1 passes through the points (–1, 3) and (11, 12). Find an equation for L1 in the form ax + by + c = 0, where a, b and c are integers

When finding the equation of a straight line there are two important figures to calculate. The first being the gradient (the slope of the line) and the second being the y intercept (where the line crosses...

Answered by Ruby B. • Maths tutor

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