Understanding Engineering Mathematics

3.1.5 Composition of functions ➤97 108➤➤ Iff(x)=x+1andg(x)= 1 x− 1 express in their simplest form: (i) f(g(x)) (ii) g(f(x)) 3.1. ...

3.2 Revision 3.2.1 Definition of a function ➤ 88 107➤ Afunctionis a relation which expresses how the value of one quantity, thed ...

Polynomial:f(x)=anxn+an− 1 xn−^1 +···+a 1 x+a 0 (allx) Reciprocal: f(x)= 1 x x= 0 Rational function – for example: f(x)= x+ 1 ( ...

y − 4 − 3 − 2 − 10 1 2 3 4 x 4 3 2 1 − 1 − 2 − 3 − 4 (1,3) (0,1) y = 2 x + 1 Figure 3.1The graph ofy= 2 x+1. Like the linear fun ...

y − 4 − 3 − 2 − 1 0 1 2 3 4 x 6 3 4 5 2 1 − 1 − 2 − 3 − 4 − 5 (−2,5) (4,5) y = x^2 − 2 x − 3 (−1,0) (3,0) (1,−4) (0,−3) (2,−3) F ...

temperatureTby theideal gas law: P= nRT V wherenRis a constant dependent on the gas. IfTis fixed then we can regard this formula ...

Examples include: f(x)= 5 x^2 , 3 x^4 +x^2 i.e. polynomials with only even powers ofx. Anodd functionchanges sign with its argum ...

We did a similar thing in completing the square (66 ➤ ). Themodulus functionf(x)=|x|is defined by (8 ➤ ) |x|=xifx≥ 0 =−xifx< ...

3.2.5 Composition of functions ➤ 89 108➤ Often we need to combine functions. For example in differentiating a function such as 1 ...

This property is another fairly obvious one for real numbers – if Jack is shorter than Jill and Jill is shorter than John then J ...

An inequality in whichf(x)is a linear (quadratic) function is called alinear (quadratic) inequality, for example: ax+b> 0 lin ...

Solution to review question 3.1.6 If 1< 4 x− 3 <4 then adding 3 throughout gives 1 + 3 < 4 x< 4 + 3 or 4< 4 x< ...

In fact we can do this generally for any value ofyby finding the inverse function off: y=f(x)= 1 x− 2 Multiplying through byx−2g ...

Here the inverse function is exactly the same as the original function. There is nothing strange about this, but it does emphasi ...

Example The generalnth degree polynomial (43 ➤ ) may be written in sigma notation as: anxn+an− 1 xn−^1 +···+a 1 x+a 0 = ∑n r= 0 ...

because all terms but the first and last cancel out. So Sn= a( 1 −rn) 1 −r Example For the series ∑^5 r= 1 1 3 r−^1 = 1 + 1 3 + ...

So, for the series 1 2 + 1 22 + 1 23 + 1 24 + 1 25 we havea=^12 andn=5, and therefore the sum is S 5 = 1 2 ( 1 − ( 1 2 ) 5 ) 1 − ...

The only placenoccurs here is in thern.Now,if|r|<1, thenrn‘tends to’ zero asn gets larger and larger, i.e.rntends to zero asn ...

integers are subtracted fromnandn−r+1 will never become zero becauseris an integer whereasnis not. So the expansion becomes an i ...

(iii) f(x)= 2 x^2 −x+1(iv)f(x)= 1 x+ 2 x=− 2 (v) f(x)= 3 x− 1 2 x+ 1 x=− 1 2 B.Iff(x)= x^2 − 1 x+ 2 obtain expressions for (i) ...