AI SL · 12.9
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Expanding, splitting fractions and increasing or decreasing
12.9.6
The letter c is a constant. Find \(\frac{dy}{dx}\) when \(y=(x-4)(x+c)\), and then its value when \(x=5\) and \(c=2\).
Have a go first. Then open the working one line at a time.
\(y=x^{2} + (c-4)x - 4c\)
Multiply out first. The rule needs a sum of terms, not a product.
\(\frac{dy}{dx}=2x - 4 + c\)
c is a constant like any number: \(cx\) gives \(c\) and \(-4c\) gives 0.
\(2(5)+2-4=8\)
Put in \(x=5\) and \(c=2\) only after you have differentiated.
Differentiate \(f(x)=\frac{3x^3+8}{2x}\), then find \(f'(2)\).
Have a go first. Then open the working one line at a time.
\(f(x)=\frac{3x^3}{2x}+\frac{8}{2x}=\frac{3}{2}x^{2} + \frac{4}{x}\)
Divide each term of the top by the bottom. Then \(\frac{4}{x}=4x^{-1}\).
\(f'(x)=3x - \frac{4}{x^{2}}\)
Multiply by the power, then take 1 off the power: \(4x^{-1}\to-4x^{-2}\).
\(f'(2)=3(2)-\frac{4}{2^2}=5\)
Put the numbers in last.
Find the values of x for which \(f(x)=4x^2-10x+7\) is decreasing.
Have a go first. Then open the working one line at a time.
\(f'(x)=8x - 10\)
A decreasing function has a negative gradient.
\(8x-10=0 \Rightarrow x=1.25\)
Solve the equation first. It gives where the gradient changes sign.
\(8x-10<0 \Rightarrow x<1.25\)
The gradient is negative to the left of the boundary, because the graph is a cup shape (positive \(x^2\) term).
Differentiate \(P=9an+\frac{12b}{n}\) with respect to n. Then find the value of \(\frac{dP}{dn}\) when \(a=2\), \(b=3\) and \(n=2\).
Have a go first. Then open the working one line at a time.
\(P=9an+12bn^{-1}\)
The variable is n. The letters a and b are constants.
\(\frac{dP}{dn}=9a - \frac{12b}{n^{2}}\)
\(9an\to9a\) and \(12bn^{-1}\to-12bn^{-2}\).
\(9(2)-\frac{12(3)}{2^2}=9\)
Put the numbers in last.
If \(4x-y=2x^2+3\), find \(\frac{dy}{dx}\), and its value when \(x=3\).
Have a go first. Then open the working one line at a time.
\(y=4x-(2x^2+3)=-2x^{2} + 4x - 3\)
Move y to one side on its own. The minus applies to both terms of the bracket.
\(\frac{dy}{dx}=-4x + 4\)
The constant −3 gives 0.
\(4-4(3)=-8\)
Put the numbers in last.
The strength S of a radio signal at distance d km from a transmitter is modelled by \(S=\frac{c}{d}\), where c is a constant. The fade rate is the rate of change of S with respect to d. (a) Find an expression for the fade rate in terms of c and d. (b) Find an expression for the fade rate in terms of S and c. (c) Station A is a quarter of the way from the transmitter to station B. Find the value of (fade rate at A) ÷ (fade rate at B).
Have a go first. Then open the working one line at a time.
\(S=cd^{-1}\Rightarrow\frac{dS}{dd}=-cd^{-2}=-\frac{c}{d^2}\)
c is a constant.
\(d=\frac{c}{S}\Rightarrow\frac{dS}{dd}=-\frac{c}{(c/S)^2}=-\frac{S^2}{c}\)
Make d the subject of \(S=\frac cd\), then substitute.
\(\text{At B: }-\frac{c}{d^2}.\ \text{At A } (\tfrac14 d):\ -\frac{c}{(d/4)^2}=-\frac{16c}{d^2}\)
Put the distance of each station into the same formula.
\(\frac{-16c/d^2}{-c/d^2}=16\)
The nearer station fades 16 times as fast.
Show that \(f(x)=2x^3+9x-5\) is increasing for all values of x.
Have a go first. Then open the working one line at a time.
\(f'(x)=6x^{2} + 9\)
Increasing means a positive gradient.
\(x^2\geq0\Rightarrow 6x^2+9\geq9\)
A square is never negative, so \(6x^2\) is never negative.
\(f'(x)>0\text{ for all }x\)
The gradient is always positive, so f is increasing for all x.