IB Math Analysis & Approaches Question bank-Topic SL 5.1 Introduction to the concept of a limit SL Paper 1

Question

Let \(f(x) = 1 + {{\text{e}}^{ – x}}\) and \(g(x) = 2x + b\), for \(x \in \mathbb{R}\), where \(b\) is a constant.

Find \((g \circ f)(x)\).

[2]
a.

Given that \(\mathop {\lim }\limits_{x \to  + \infty } (g \circ f)(x) =  – 3\), find the value of \(b\).

[4]
b.
Answer/Explanation

Markscheme

attempt to form composite     (M1)

eg\(\,\,\,\,\,\)\(g(1 + {{\text{e}}^{ – x}})\)

correct function     A1     N2

eg\(\,\,\,\,\,\)\((g \circ f)(x) = 2 + b + 2{{\text{e}}^{ – x}},{\text{ }}2(1 + {{\text{e}}^{ – x}}) + b\)

[2 marks]

a.

evidence of \(\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ – x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ – x}})\)     (M1)

eg\(\,\,\,\,\,\)\(2 + b + 2{{\text{e}}^{ – \infty }}\), graph with horizontal asymptote when \(x \to \infty \)

Note:     Award M0 if candidate clearly has incorrect limit, such as \(x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}\).

evidence that \({{\text{e}}^{ – x}} \to 0\) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ – x}}) = 0,{\text{ }}1 + {{\text{e}}^{ – x}} \to 1,{\text{ }}2(1) + b =  – 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0\), graph of \(y = {{\text{e}}^{ – x}}\) or

\(y = 2{{\text{e}}^{ – x}}\) with asymptote \(y = 0\), graph of composite function with asymptote \(y =  – 3\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2 + b =  – 3\)

\(b =  – 5\)     A1     N2

[4 marks]

b.

MAA SL 5.1-5.3 DERIVATIVES – TANGENT AND NORMAL [concise]-neha

Question

[without GDC]

Differentiate the following functions:

Answer/Explanation

Ans

Derivatives are given in the 2nd column.

Question

[without GDC]

Let \(f(x)=2x^{3}+\ln x\)

(a)  Find \(f'(x)\).

(b) Find the gradient of the curve \(y=f(x)\) at \(x=1\).

Answer/Explanation

Ans

(a)  \(f'(x)=6x^{2}+\frac{1}{x}\)                     (b)    \(f'(1)=6+1=7\)

Question

[with GDC]

Let \(f(x)=\frac{x^{3}+1}{\sin x}\)

(a)    Find \(f'(x)\).

(b)    Find the gradient of the curve \(y=f(x)\)

(ⅰ)   at \(x=\frac{\pi }{4}\)              (ⅰⅰ)   at \(x=1 rad\).

Answer/Explanation

Ans

(a)   \(f'(x)=\frac{3x^{2}\sin x-(x^{3}+1)\cos x}{\sin ^{2}x}\)

(b)    Directly by GDC   (i)  \(f'(\frac{\pi }{4})\cong 0.518\)       (ii)   \(f'(1)\cong 2.04\)

[Notice: the exact value for (i) is \(f'(\frac{\pi }{4})=\frac{3\pi ^{2}}{16}\sqrt{2}-\frac{\pi ^{3}+64}{64}\sqrt{2}]\)

Question

[without GDC]

Given the following values at \(x =1\)

Calculate the derivatives of the following functions at \(x =1\)

(ⅰ)   \(y=2f(x)+3g(x)\)                                (ⅰⅰ)    \(y=f(x)g(x)\)

(ⅰⅰⅰ)   \(y=\frac{f(x)}{g(x)}\)                           (iv)   \(y=2x^{3}+1+5f(x)\)

Answer/Explanation

Ans

(a)   \(\frac{\mathrm{d} y}{\mathrm{d} x}=2f'(x)+3g'(x)\),  at \(x=1\)  the value is \(23\)

(b)   \(\frac{\mathrm{d} y}{\mathrm{d} x}=f'(x)g(x)+f(x)g'(x)\), at \(x=1\) the value is \(22\)

(c)    \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}\), at \(x=1\) the value is \(\frac{2}{9}\)

(d)    \(\frac{\mathrm{d} y}{\mathrm{d} x}=6x^{2}+5f'(x)\), at \(x=1\) the value is \(26\)

Question

[without GDC]

Let \(f(x)=x^{3}-2x^{2}-1\).

(a)   Find \(f'(x)\)

(b)   Find the gradient of the curve of \(f(x)\) at the point \((2,-1)\).

Answer/Explanation

Ans

(a)    \(f'(x)=3x^{2}-4x-0=3x^{2}-4x\)

(b)    Gradient = \(f'(2)=3\times 4-4\times 2=4\)

Question

[without GDC]

Let \(f(x)=6\sqrt[3]{x^{2}}\).  Find   \(f'(x)\).

Answer/Explanation

Ans

\(f(x)=6x^{\frac{2}{3}}\)

\(f'(x)=4x^{\frac{1}{3}}\left ( \frac{4}{x^{\frac{1}{3}}}=\frac{4}{\sqrt[3]{x}} \right )\)

Question

[without GDC]

Let \(h(x)=\frac{6x}{\cos x}\).  Find   \(h'(0)\)

Answer/Explanation

Ans

METHOD 1 (quotient)

\(h'(x)=\frac{(\cos x)(6)-(6x)(-\sin x)}{(\cos x)^{2}}\)

\(h'(0)=\frac{(\cos 0)(6)-(6\times 0)(-\sin 0)}{(\cos 0)^{2}}=6\)

METHOD 2 (product)

\(h(x)=6x\times (\cos x)^{-1}\)

\(h'(x)=(6x)(-(\cos x)^{-2}(-\sin x))+(6)(\cos x)^{-1}\)

\(h'(0)=(6\times 0)(-(\cos 0)^{-2}(-\sin 0))+(6)(\cos 0)^{-1}=6\)

Question

[without GDC]

Let \(g(x)=2x \sin x\).

(a)  Find  \(g'(x)\).

(b)  Find the gradient of the graph of \(g\) at \(x=\pi\).

Answer/Explanation

Ans

(a)  \(g'(x)=2\sin x+2\ x \cos x\)

(b)  \(g'(\pi )=2\sin \pi +2\pi \cos \pi=-2\pi\)

Question

[without GDC]

Given the function \(f(x)=x^{2}-3bx+(c+2)\), determine the values of \(b\) and \(c\) such that

\(f(1)=0\) and \(f'(3)=0\).

Answer/Explanation

Ans

\(f(1)=1^{2}-3b+c+2=0\)

\(f'(x)=2x-3b\),

\(f'(3)=6-3b=0\Rightarrow 3b=6\Rightarrow b=2\)

\(1-3(2)+c+2=0\Rightarrow c=3\)

Question

[without GDC]

Consider the function \(f(x)=k\sin x+3x\), where \(k\) is a constant.

(a)  Find \(f'(x)\).

(b)  When \(x=\frac{\pi }{3}\), the gradient of the curve of \(f(x)\) is \(8\). Find the value of \(k\).

Answer/Explanation

Ans

(a)   \(f'(x)=k\cos x+3\)

(b)    \(k\cos \left ( \frac{\pi }{3} \right )+3=8\Rightarrow k\left ( \frac{1}{2} \right )+3=8\Rightarrow k=10\)

Question

[without GDC]

 Let \(f(x)=\frac{3x^{2}}{5x-1}\)

(a)  Write down the equation of the vertical asymptote of  \(y=f(x)\).

(b)   Find  \(f'(x)\). Give your answer in the form \(\frac{ax^{2}+bx}{(5x-1)^{2}}\) where \(a\) and \(b \, \epsilon\, \mathbb{Z}\).

Answer/Explanation

Ans

(a)   \(x=\frac{1}{5}\)

(b)    \(f'(x)=\frac{(5x-1)(6x)-(3x^{2})(5)}{(5x-1)^{2}}=\frac{30x^{2}-6x-15x^{2}}{(5x-1)^{2}}=\frac{15x^{2}-6x}{(5x-1)^{2}}\)

Question

[without GDC]

The graph of the function \(y=f(x), 0\leq x\leq 4\), is shown below.

(a)  Write down the value of         (i) \(f (1)\)            (ii) \(f (3)\)

(b)  Write down the value of         (i) \(f ‘(1)\)            (ii) \(f ‘(3)\)

Answer/Explanation

Ans

(a)        (i) \(1\)                (ii) \(0.5\)

(b)        (i) \(0\)               (ii) \(-\frac{1}{2}\)

Question

[without GDC]

Part of the graph of the periodic function \(f\) is shown below. The domain of \(f\) is \(0\leq x\leq 15\) and the period is \(3\).

(a)    Find   (i) \(f(2)\)                 (ii)  \(f'(6.5)\)                      (iii) \(f'(14)\)

(b)    How many solutions are there to the equation \(f(x)=1\) over the given domain?

Answer/Explanation

Ans

(a)      (i)  \(1\)

(ii) \(2\)

(iii)  \(f'(14)=f'(2)(or f'(5) or f'(8))=-1\)

(b)     There are five repeated periods of the graph, each with two solutions, i.e number of solutions is \(5\times 2=10\)

Question

[with GDC]

Let  \(f(x)=x\cos x, for 0\leq x\leq 6\).

(a) Find \(f'(x)\).

(b) On the grid below, sketch the graph of \(y=f'(x)\).

(c) Write down the range of the function \(y=f'(x)\), for \(0\leq x\leq 6\)

Answer/Explanation

Ans

(a) \(f ′(x) = \cos x – x \sin x\)

(b)         

(c)     \(\ y \epsilon [-2.38, 5.10]\)

Question

[without GDC]

Let \(f(x)=2x^{2}-12x+10\)

(a)  Find the tangent line and the normal line at \(x=2\)

(b)  Find the tangent line and the normal line at \(x=1\)

(c)  Find the tangent line and the normal line at \(x=3\)

Answer/Explanation

Ans

(a)      Tangent line:    \(y+6=-4(x-2)\)               (i.e. \(y=-4x+2\))

Normal line:    \(y+6=\frac{1}{4}\)        (i.e. \(y=\frac{1}{4}x-\frac{13}{2}\)

(b)      Tangent line:    \(y=-8(x-1)\)                    (i.e. \(y=-8x+8\))

Normal line:     \(y=\frac{1}{8}(x-1)\)   (i.e. \(y=\frac{1}{8}x-\frac{1}{8}\))

(c)     Tangent line:    \(y=-8\) ,           Normal line: \(x = 3\)

Question

[without GDC]

Let \(f(x)=2x^{2}-12x+10\)

(a) Find the tangent line which is parallel to the line \(y=4x-7\)

(b) Find the tangent line which is perpendicular to the line \(y=\frac{1}{4}x-7\)

Answer/Explanation

Ans

(a) \(y = 4x-22\)                 (b) \(y=-4x+2\)

Question

[without GDC]

Find the equation of the tangent line and the equation of the normal to the curve with equation \(y=x^{3}+1\) at point \((1,2)\).

Answer/Explanation

Ans

\(y=x^{3}+1 \frac{\mathrm{d} y}{\mathrm{d} x}=3x^{2}\)

Therefore at point where \(x = 1, slope = 3\)

Equation of tangent: \(y – 2 = 3(x – 1)\Rightarrow y = 3x – 1\)

Slope of normal =\(-\frac{1}{3}\)

Equation of normal: \(y – 2 =-\frac{1}{3}(x-1)\Rightarrow 3y-6=-x+1\Rightarrow x+3y-7=0\) OR  \(y=-\frac{1}{3}x+2\frac{1}{3}\)

Question

[without GDC]

Consider the function \(f(x)=4x^{3}+2x\). Find the equation of the normal to the curve of \(f\) at the point where \(x=1\).

Answer/Explanation

Ans

\(f'(x)= 12x^{2}+2\)

When \(x = 1, f(1)=6\)

When \(x = 1, f'(1)=14\)

Equation is \(y-6=-\frac{1}{14}(x-1)  \left ( y=-\frac{1}{14} x+\frac{85}{14}, y=-0.0714x+6.07\right )\)

Question

[without GDC]

Find the coordinates of the point on the graph of \(y=x^{2}-x\) at which the tangent is parallel to the line \(y=5x\) .

Answer/Explanation

Ans

\(y = x^{2}-x \frac{\mathrm{d} y}{\mathrm{d} x}=2x-1\).

Line parallel to \(y = 5x \Rightarrow 2x-1 = 5\Rightarrow x = 3\) so \(y = 6\)

Question

[without GDC]

Let \(f(x)=kx^{4}\). The point P\((1,k)\) lies on the curve of \(f\). At P, the normal to the curve is parallel to \(y=-\frac{1}{8}x\). Find the value of \(k\).

Answer/Explanation

Ans

gradient of tangent = \(8\)

\(f'(x)=4kx^{3}\)

\(4kx^{3} = 8 \Rightarrow kx^{3} = 2\)

substituting \(x=1, k=2\)

Question

[without GDC]

Consider the function \(f:x \mapsto 3x^2 -5x+k\).

The equation of the tangent to the graph of \(f\) at \(x=p\) is \(y=7x-9\) .

(a) Write down \(f'(x)\).

(b) Find the value of      (i) \(p\) ;          (ii) \(k\) .

Answer/Explanation

Ans

(a)    \(f'(x)= 6x-5\)

(b)    \(f'(p) = 7 \Rightarrow 6p-5=7 \Rightarrow p=2\)

(c)      Setting  \(y(2) = f(2)\)

Substituting  \(y(2)=7\times 2-9=5\),  and \(f(2)=3\times 2^{2}-5\times 2+k=k+2\)

\(k+2=5\Rightarrow k=3\)

Question

[without GDC]

Let \(f(x)=e^{x} \cos x\).

(a)  Find the gradient of the normal to the curve of \(f\) at \(x= \pi\).

(b)  Find the gradient of the tangent to the curve of \(f\) at \(x= \frac{\pi}{4}\).

Answer/Explanation

Ans

(a)     \(f'(x)=e^{x}\times (-\sin x)+ \cos x\times e^{x}=e^{x}\cos x-e^{x}\sin x\)

\(f'(x)=e^{\pi }\cos \pi -e^{\pi }\sin \pi=-e^{\pi }\)

gradient of normal = \(\frac{1}{e^{x}}\)

(b)     \(f’\left ( \frac{\pi }{4} \right )=0\)

Question

[with / without GDC]

Consider the curve with equation \(f(x)=px^{2}+qx\), where \(p\) and \(q\) are constants.

The point A\((1, 3)\) lies on the curve. The tangent to the curve at A has gradient \(8\).

(a)  Find the value of \(p\) and of \(q\) .

(b)  Find the equations of the tangent line and the normal at \(x=0.2\).

Answer/Explanation

Ans

(a)     \(f(1)=3 \Rightarrow p+q=3\)

\(f'(x)=2px+q\)

\(f'(1)=8\Rightarrow 2p+q=8\)

\(p=5,   q=-2\)

(b)     \(f'(x)=10x-2\)

\(f'(0.2)=0, at x = 0.2 y = -0.2\)

Tangent \(y = – 0.2\) (horizontal line)

Normal \(x = 0.2\) (vertical line)

Question

[without GDC]

Consider the function

(a)     Find the equation of the tangent to the graph of \(h\) at the point where \(x=a\),

\((a\neq 0)\). Write the equation in the form \(y=mx+c\).

(b)    Show that this tangent intersects the \(x\) -axis at the point \((-4a,0)\).

Answer/Explanation

Ans

(a)    At \(x=a, h(x)=a^{\frac{1}{5}}\)

\(h'(x)=\frac{1}{5}x\frac{-4}{5} \Rightarrow h'(a)= \frac{1}{5a^{\frac{4}{5}}}\)= gradient of tangent

\(\Rightarrow y-a\frac{1}{5}=\frac{1}{5a^{\frac{4}{5}}}(x-a)=\frac{1}{5a^{\frac{4}{5}}}x-\frac{1}{5}a^{\frac{1}{5}}\Rightarrow y=\frac{1}{5a^{\frac{4}{5}}}x+\frac{4}{5}a\frac{1}{5}\)

(b)     tangent intersects \(x\)-axis \(\Rightarrow y=0 \Rightarrow \frac{1}{5a^{\frac{4}{5}}}x=-\frac{4}{5}a^{\frac{1}{5}}\)

\(\Rightarrow x=5a^{\frac{4}{5}}\left ( -\frac{4}{5}a^{\frac{1}{5}} \right )=-4a\)

Question

[without GDC]

The following diagram shows part of the graph of a quadratic function, with equation in the form

\(y=(x-p)(x-q), where p, q \, \epsilon \ Z\).

(a)      (i)   Write down the value of \(p\) and of \(q\)

(ii) Write down the equation of the axis of symmetry of the curve.

(b)      Find the equation of the function in the form \(y=(x-h)^{2}+k, where h, k\, \epsilon \ Z.\)

(c)       Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\)

(d)       Let \(T\) be the tangent to the curve at the point \((0, 5)\). Find the equation of \(T\).

Answer/Explanation

Ans

(a)      (i) \(p = 1, q = 5\) (or \(p = 5, q = 1\))

(ii) \(x = 3\) (must be an equation)

(b)      \(y = (x-1)(x-5)=x^{2}-6x+5=(x-3)^{2}-4 (h=3, k=-4)\)

(c)      \(\frac{\mathrm{d} y}{\mathrm{d} x}=2(x-3) (=2x-6)\)

(d)      When \(x=0, \frac{\mathrm{d} y}{\mathrm{d} x}=-6\)

\(y-5=-6(x-0) (y=-6x+5\) or equivalent)

Question

[with GDC]

The function \(f(x)\) is defined as \(f(x)=-x(x-h)^{2}+k\). The diagram below shows part of the graph of \(f(x)\). The maximum point on the curve is P\((3,2)\).

(a)    Write down the value of    (i) \(h\)      (ii) \(k\)

(b)    Show that \(f(x)\) can be written as \(f(x)=-x^{2}+6x-7\).

(c)    Find \(f'(x)\).

The point Q lies on the curve and has coordinates \((4, 1)\). A straight line \(L\), through Q, is perpendicular to the tangent at Q.

(d)     (i)  Find the equation of \(L\).

(ii)  The line \(L\) intersects the curve again at R. Find the \(x\)-coordinate of R.

Answer/Explanation

Ans

(a)    \(h=3\)          \(k=2\)

(b)    \(f(x)= -x(x-3)^{2}+2=-x^{2}+6x-9+2 = -x^{2}+6x-7\)

(c)    \(f'(x)=-2x+6\)

(d)    (i)       tangent gradient = \(-2\) gradient of \(L=\frac{1}{2}\)

\(y-1=\frac{1}{2}(x-4)\Rightarrow y=\frac{1}{2}x-1\)

(ii)      \(-x^{2}+6x-7=\frac{1}{2}x-1\Leftrightarrow 2x^{2}-11x+12=0\Leftrightarrow x=1.5\) or \(x=4\) so \(x=1.5\)

OR by GDC  \(-x^{2}+6x-7=\frac{1}{2}x-1\Rightarrow x=1.5\)

Question

[with GDC]

The function \(f\) is defined by \(f:x \mapsto -0.5x^{2}+2x+2.5.\)

(a)  Write down              (i) \(f'(x)\);                       (ii) \(f ‘(0)\).

(b) Let \(N\) be the normal to the curve at the point where the graph intercepts the \(y\)-axis. Show that the equation of \(N\) may be written as \(y=-0.5x+2.5\).

Let  \(g:x \mapsto -0.5x+2.5\)

(c)      (i) Find the solutions of \(f(x)= g(x)\)

(ii) Hence find the coordinates of the other point of intersection of the normal and the curve.

Answer/Explanation

Ans

(a)     (i)    \(f'(x)=-x+2\)

(ii)   \(f'(0)=2\)

(b)    Gradient of tangent at \(y-intercept\) = \(f'(0)=2\)

gradient of normal =  \(y-2.5 =-0.5(x-0)\Rightarrow y=-0.5x+2.5\)

(c)      (i)    \(-0.5x^{2}+2x+2.5=-0.5x+2.5\Rightarrow x=0\) or \(x=5\)

(ii)     Curve and normal intersect when \(x = 0\) or \(x = 5\)

Other point is when \(x = 5 \Rightarrow y = –0.5(5) + 2.5 = 0\) (so other point \((5, 0)\)

Question

[with GDC]

The equation of a curve may be written in the form \(y=a(x-p)(x-q)\). The curve intersects the \(x\)-axis at A\((–2, 0)\) and B\((4, 0)\). The curve of \(y= f(x)\) is shown in the diagram below.

(a)     (i)  Write down the value of \(p\) and of \(q\).

(ii)  Given that the point \((6, 8)\) is on the curve, find the value of \(a\).

(iii)  Write the equation of the curve in the form \(y= ax^{2}+bx+c\).

(b)  A tangent is drawn to the curve at a point \(P\). The gradient of this tangent is \(7\). Find the coordinates of \(P\).

(c)  The line \(L\) passes through B\((4, 0)\), and is normal to the curve at B.

(i) Find the equation of \(L\).

(ii) Find the \(x\)-coordinate of the point where \(L\) intersects the curve again.

Answer/Explanation

Ans

(a)     (i)     \(p=-2 q=4\) (or \(p=4, q=-2)\)

(ii)    \(y=a(x+2)(x+4)\Leftrightarrow 8=a(6+2)(6-4)\Leftrightarrow 8=16a \Leftrightarrow a=\frac{1}{2}\)

(iii)   \(y=\frac{1}{2}(x+2)(x-4)\Rightarrow y=\frac{1}{2}(x^{2}-2x-8)\Rightarrow y=\frac{1}{2}x^{2}-x-4\)

(b)     \(\frac{\mathrm{d} y}{\mathrm{d} x}=x-1\)

\(x-1=7\Leftrightarrow x=8, y=20\) (P is \((8,20))\)

(c)     (i)       when \(x=4,m_{T}=4-1=3 \Rightarrow m_{N} = -\frac{1}{3}\)

\(y-0=-\frac{1}{3}(x-4)\)           \(\left ( y=-\frac{1}{3}x+\frac{4}{3} \right )\)

(ii)     \(\frac{1}{2}x^{2}-x-4=-\frac{1}{3}x+\frac{4}{3}\Leftrightarrow x=-\frac{8}{3}\) or  \(x=4\)

\(x = -\frac{8}{3} (-2.67)\)

Question

[without GDC]

The function \(f\) is given by \(f(x)=\frac{2x+1}{x-3}, x\, \epsilon \ R, x\neq 3\)

(a)     (i) Show that \(y=2\) is an asymptote of the graph of \(y= f(x)\) .

(ii) Find the vertical asymptote of the graph.

(iii) Write down the coordinates of the point \(P\) at which the asymptotes intersect.

(b)      Find the points of intersection of the graph and the axes.

(c)       Hence sketch the graph of \(y= f(x)\), showing the asymptotes by dotted lines.

(d)      Show that \(f'(x)= \frac{-7}{(x-3)^{2}}\) and hence find the equation of the tangent at the point \(S\) where \(x=4\) .

(e)       The tangent at the point \(T\) on the graph is parallel to the tangent at \(S\). Find the coordinates of \(T\).

(f)       Show that \(P\) is the midpoint of \([ST]\).

Answer/Explanation

Ans

(a)     (i)    \(f(x)=\frac{2x+1}{x-3}=2+\frac{7}{x-3} OR f(x)=\frac{2+\frac{1}{x}}{1-\frac{3}{x}}\)

Therefore as \(\left | x \right |\rightarrow \infty f(x)\rightarrow 2\Rightarrow y=2\) is an asymptote

Note: inexact methods based on the ratio of the coefficients of x also accepted

(ii)    Asymptote at \(x = 3\)

(iii)    \(P(3, 2)\)

(b)     \(f(x)=0\Rightarrow x=-\frac{1}{2} \left ( -\frac{1}{2},0 \right )\)

\(x=0\Rightarrow f(x)=-\frac{1}{3} \left ( 0,-\frac{1}{3} \right )\)

(c)                                   

(d)     \(f'(x)=\frac{(x-3)(2)-(2x+1)}{(x-3)^{2}}=-\frac{7}{(x-3)^{2}}\)

Therefore slope when \(x = 4\) is \(–7\)

And \(f (4) = 9\)          ie \(S(4, 9)\)

Equation of tangent: \(y – 9 = –7(x – 4) \Rightarrow 7x + y – 37 = 0\)

(e)     at T, \(\frac{-7}{(x-3)^{2}}=-7\Rightarrow (x-3)^{2}=1\Rightarrow x-3=\pm 1\)

(f)      Midpoint \(\ [ST]= \left ( \frac{4+2}{2},\frac{9-5}{2} \right )=(3,2)\) = point \(P\)

Question

[with GDC]

Let \(f(x)= x^{3}-3x^{2}-24x+1\)

(a)  Find   \(f'(x)\).

The tangents to the curve of \(f\) at the points \(P\) and \(Q\) are parallel to the \(x\) – axis, where P is to the left of Q.

(b)  Calculate the coordinates of P and of Q.

Let \(N_{1}\) and \(N_{2}\) be the normals to the curve at P and Q respectively.

(b) Write down the coordinates of the points where

(i) the tangent at P intersects \(N_{2}\);

(ii) the tangent at Q intersects \(N_{1}\).

Answer/Explanation

Ans

(a)     \(f'(x)=3x^{2}-6x-24\)

(b)     Tangents parallel to the \(x\)-axis mean maximum and minimum (see graph) EITHER by GDC P\((-2, 29)\) and Q\((4, -79)\)

OR \(f'(x)=0\Leftrightarrow 3x^{2}-6x-24=0\Leftrightarrow\) x=\(-2\) or x=\(4\)

Coordinates are P\((-2, 29)\) and Q\((4, -79)\)

(c)                               

(i) \((4, 29)\)                     (ii) \((-2, -79)\)

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