Top answers

Maths
IB

The fifth term of an arithmetic sequence is equal to 6 and the sum of the first 12 terms is 45. Find the first term and the common difference.

Arithmatic term n, Un= U1+(n-1)d. Where U1 is the first term of the sequence and d is the common difference. U5=U1+4d=6. U1=6-4d. Sum of arithmatic terms up to term n, Sn=n/2(2U1+(n-1)d). S12=12/2(2(6-4d)...

Answered by Jasmin S. Maths tutor
8371 Views

Consider the functions f and g where f(x)=3x-5 and g(x)=x-2. (a) Find the inverse function for f. (b) Given that the inverse of g is x+2, find (g-1 o f)(x).

(a) In order to find the inverse of a function, it is easiest to swap x and y and solve for y. Here this would give, x=3y-5 => x+5=3y => (x+5)/3=y. Hence, f-1(x)=(x+5)/3. (b) Here it is important to...

Answered by Rebecca M. Maths tutor
2544 Views

Find out the stationary points of the function f(x)=x^2*e^(-2x)

Using the product rule (u'v+v'u, where u and v are the chosen substitutes) to find the first derivative will be dy/dx=x'=2xe^(-2x)+x^(2)e^(-2x)(-2)=2xe^(-2x)(x-x^2). This will give the d...

Answered by Bilkan I. Maths tutor
2798 Views

Find the constant term in the binomial expansion of (3x + 2/(x^2))^33

Recall that the general term in the binomial expansion of (x+y)^n is (nCr)(x^n-r)(y^r), so by the binomial theorem, the entire expansion is the sum of these terms from r = 0 to n. In this case, n = 33, th...

Answered by Elias G. Maths tutor
6504 Views

If f(x)=(x^3−2x)^5 , find f'(x).

f(x)=(x3-2x)5
If we look at this function, we can see that it can be split into two functions, one hiding in the other one. Because of that, to solve this problem we will need t...

Answered by Radovan S. Maths tutor
4905 Views

We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences