# 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.

## 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:

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 $$x=1$$.

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 $$x=1 rad$$.

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)   $$y=2x^{3}+1+5f(x)$$

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 $$(2,-1)$$.

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   $$f'(x)$$.

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   $$h'(0)$$

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 $$x=\pi$$.

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 $$f'(3)=0$$.

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 $$k$$.

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 $$b \, \epsilon\, \mathbb{Z}$$.

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) $$f ‘(3)$$

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?

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 $$0\leq x\leq 6$$

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 $$x=3$$

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 $$y=\frac{1}{4}x-7$$

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 $$(1,2)$$.

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 $$x=1$$.

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 $$y=5x$$ .

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 $$k$$.

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) $$k$$ .

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 $$x= \frac{\pi}{4}$$.

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 $$x=0.2$$.

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 $$(-4a,0)$$.

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 $$T$$.

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.

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.

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.

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 $$[ST]$$.

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 $$N_{1}$$.

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) $$(-2, -79)$$