Partial Fraction Questions

(a)

Multiplying the given by the denominator of the left-hand side gives,

By substituting values of $x$ into this to eliminate terms, we can find the values of the unknowns (See Section 1.3 page 9).

To find $B$ we need to have $x-3=0$, so $x=3$. Substituting this value gives

So

$$B=4$$

$C$ can be found in a similar fashion with $1-2x=0$ giving $x=\frac{1}{2}$. Substituting this in gives

So

$$C=-2$$

$A$ cannot be found by substitution as no value of $x$ will leave only $A$. To find $A$ we need to compare coefficients.

So, substituting what we know, multiplying out brackets and simplifying

Equating any of the coefficients we get $A=3$ (See Section 1.3 page 9)

For example, comparing the $x$ term coefficients, we get $7A-10=11$ giving $A=3$

---

(b)

Collecting all from (a) together

This is presented better thus

To show that a function is decreasing we need to show that the derivative of the function is negative for all of the domain.

So, we need to find the derivative of $f\left(x\right)$

The first term is easy as the derivative of a constant is zero. The second and third are like one another.

Let $y=4{(x-3)}^{-1}$, then using the chain rule (see section 9.3 page 237)

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$

Here $u=x-3$ and $y=4u^{-1}$ so

$$\frac{du}{dx}=1$$ $$\frac{dy}{du}=-4u^{-2}$$

So,

$$\frac{dy}{dx}=-4u^{-2}\times 1$$

Back substituting and simplifying

$$\frac{dy}{dx}=-4{(x-3)}^{-2}$$

A similar approach gives the third term derivative which is given here along with the complete derivative of $f\left(x\right)$.

$$f^{\text{'}}(x)=-4{(x-3)}^{-2}-4{(1-2x)}^{-2}$$

And as fractions

$$f^{\text{'}}(x)=-\frac{4}{{(x-3)}^2}-\frac{4}{{(1-2x)}^2}$$

Now, for all $x>3$ both the denominators will be positive, resulting in each fraction being positive and the whole sum negative. This proves that $f^{\text{'}}\left(x\right)<0$ for all $x>3$, so, the function is decreasing over the domain.

Question 2:

TBC