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A curve has equation y = f(x) and passes through the point (4, 22). Given that f ′(x) = 3x^2 – 3x^(1/2) – 7, use integration to find f(x), giving each term in its simplest form.

Firstly we can use the difference rule to split f'(x) into three components which we can consider separately. Then using the knowledge that the integral of x^n is 1/(n+1)*x^(n+1) we get the expression for f(...
AS
Answered by Abbey S. • Maths tutor
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y = 4x/(x^2+5). a) Find dy/dx, writing your answer as a single fraction in its simplest form. b) Hence find the set of values of x for which dy/dx < 0

a) We need to differentiate this equation using the quotient rule (Given that it is a fraction with an x term on both the top and bottom of the fraction). We assign the numerator and denominator as follows: ...
JF
Answered by James F. • Maths tutor
8467 Views

A curve has parametric equations: x=(t-1)^3 and y= 3t - 8/(t^2). Find dy/dx in terms of t. Then find the equation of the normal at the point on the curve where t=2.

dx/dt = 3(t-1) 2 dy/dt = 3 + 16t -3 dy/dx=(dy/dt)(dt/dx) dy/dx = 3 + 16t -3 / 3(t-1) 2 At t=2 dy/dx= (3 + 16/8) / 3 = 5/3 Gradient of the normal = -3/5with t=2 y-4=0x-1=0 y=mx + c y - 4 = -3/5(x-1) 3x +5y -2...
JH
Answered by Jasmin H. • Maths tutor
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How can functions be transformed?

A function, y = f(x), with y on the vertical axis and x on the horizontal axis, can be transformed by 3 different ways: It can be stretched (or shrunk)If y = f(ax), the function is stretched by a scale facto...
JM
Answered by Jack M. • Maths tutor
3525 Views

The second and fifth terms of a geometric series are 750 and -6 respectively. Find: (1) the common ratio; (2) the first term of the series; (3) the sum to infinity of the series

x n = ar (n-1) (1) x 2 = 750 = ar 1 (2) x 5 = -6 = ar 4 divide second equation by first-6/750 = r 3 r 3 = -0.008r= -0.2Insert into first equation.750 = a * -0.2a = -3750Sum to infinite series = a(1/(1-r))(in...
HP
Answered by Henry P. • Maths tutor
6691 Views