Introduction to calculus

The graph of the function $y=f(x)$, $0\leq x\leq4$, 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) $f'(3)$

Part of the graph of the periodic function $f$ is shown below. The domain of $f$ is $0\leq x\leq15$ 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?

Part of the graph of $f(x)=\dfrac{e^x-1}{x}$ is shown below. The function is not defined at $x=0$.

(a) Complete the values of $f$, correct to 4 decimal places, on the tables below:

For positive values of $x$ near 0 (i.e. $0^+$)

$x$	$f(x)$
$0.1$
$0.01$
$0.001$

For negative values of $x$ near 0 (i.e. $0^-$)

$x$	$f(x)$
$-0.1$
$-0.01$
$-0.001$

[4]

(b) Deduce the value of the limit: $\displaystyle \lim_{x\to0}\dfrac{e^x-1}{x}$[1]

Part of the graph of $f(x)=\dfrac{e^{x-2}-x+1}{(x-2)^2}$ is shown below:

(a) Write down, correct to 4 s.f., the value of $f(0)$[1]

(b) Explain why $f(2)$ is not defined.[1]

(c) Complete the values of $f$, correct to 4 s.f. on the tables below:

For values of $x$ near $2^+$:

$x$	$f(x)$
$2.1$
$2.01$
$2.001$

For values of $x$ near $2^-$

$x$	$f(x)$
$1.9$
$1.99$
$1.999$

[4]

(d) Deduce the value of the limit: $\displaystyle \lim_{x\to2}\dfrac{e^{x-2}-x+1}{(x-2)^2}$[1]

Let $P=\dfrac{4Q-4\ln(1+Q)}{Q^2}$, $Q\neq0$, $-0.5\leq Q\leq4$,

(a) Write down the values of P for

(i) $Q=-0.5$

(ii) $Q=1$,

(iii) $Q=4$.[3]

(b) On the following diagram, sketch the graph of P vs Q, for $-0.5\leq Q\leq4$[2]

(c) P is not defined for $Q=0$. By investigating the values of P corresponding to values of Q near zero, deduce the value of $\displaystyle \lim_{Q\to0}P$.[2]

Let $f(x)=\dfrac{3x+2}{x+5}$

(a) Complete the values of $f$, correct to 6 s.f. on the table below:

$x$	$f(x)$
$100$
$1\ 000$
$1\ 000\ 000$

[2]

(b) Deduce the value of the limit: $\displaystyle \lim_{x\to\infty}\dfrac{3x+2}{x+5}$[1]

(c) Complete the values of $f$, correct to 6 s.f. on the table below:

$x$	$f(x)$
$-100$
$-1\ 000$
$-1\ 000\ 000$

[2]

(d) Deduce the value of the limit: $\displaystyle \lim_{x\to-\infty}\dfrac{3x+2}{x+5}$.[1]

(e) By using a similar rationale deduce the value of the limits: $\displaystyle \lim_{x\to\pm\infty}\dfrac{3x+2}{2x+5}$.[2]

(a) By observing the graph of the function $f(x)=\dfrac{x+3}{x-2}$ on your GDC, or otherwise find the values of the following limits

(i) $\displaystyle \lim_{x\to3}\dfrac{x+3}{x-2}$

(ii) $\displaystyle \lim_{x\to+\infty}\dfrac{x+3}{x-2}$

(iii) $\displaystyle \lim_{x\to-\infty}\dfrac{x+3}{x-2}$[3]

(b) Investigate whether each of the following side limits is $+\infty$ or $-\infty$:

(i) $\displaystyle \lim_{x\to2^+}\dfrac{x+3}{x-2}$

(ii) $\displaystyle \lim_{x\to2^-}\dfrac{x+3}{x-2}$

(iii) $\displaystyle \lim_{x\to2^+}\dfrac{x-3}{x-2}$

(iv) $\displaystyle \lim_{x\to2^-}\dfrac{x-3}{x-2}$[4]

The gradient of the curve $f(x)=x^2$ at point $x=a$ is defined by the limit

$$m_a=\lim_{h\to0}\frac{(a+h)^2-a^2}{h}$$

For example, the gradient at $x=1$ is $m_1=2$ since $\displaystyle \lim_{h\to0}\frac{(1+h)^2-1^2}{h}=2$.

(a) By using the graph mode on your GDC, find

(i) $\displaystyle \lim_{h\to0}\frac{(2+h)^2-2^2}{h}$ and hence the gradient $m_2$.

(ii) $\displaystyle \lim_{h\to0}\frac{(3+h)^2-3^2}{h}$ and hence the gradient $m_3$.[4]

(b) Find the gradient of the curve

(i) at $x=5$. (ii) $x=-5$[3]

(c) Deduce the value of the gradient $m_a$ in terms of $a$ in general.[1]

The gradient of the curve $f(x)=\ln x$ at point $x=a$ is defined by the limit

$$m_a=\lim_{h\to0}\frac{\ln(a+h)-\ln a}{h}$$

For example, the gradient at $x=1$ is $m_1=1$ since $\displaystyle \lim_{h\to0}\frac{\ln(1+h)-\ln1}{h}=1$.

(a) By using the graph mode on your GDC, find

(i) $\displaystyle \lim_{h\to0}\frac{\ln(2+h)-\ln2}{h}$ and hence the gradient $m_2$.

(ii) $\displaystyle \lim_{h\to0}\frac{\ln(5+h)-\ln5}{h}$ and hence the gradient $m_5$.[4]

(b) Find the gradient $m_{10}$ of the curve at $x=10$.[2]

(c) Deduce the value of the gradient $m_a$ in terms of $a$ in general.[2]

(a) Write down the value of $e^2$ correct to 5 s.f.[1]

(b) Write down, correct to 5 s.f., the values of the expression $\left(1+\dfrac{2}{n}\right)^n$ for the values of $n$ shown on the following table:

$n$	$\left(1+\dfrac{2}{n}\right)^n$
$1$
$10$
$100$
$1\ 000$
$1\ 000\ 000$

[3]

(c) Hence, guess the value of $\displaystyle \lim_{n\to+\infty}\left(1+\dfrac{2}{n}\right)^n$[1]

(d) By following a similar rationale guess the value of $\displaystyle \lim_{n\to+\infty}\left(1+\dfrac{3}{n}\right)^n$[3]

The graph of the function $f$ is shown below

(a) Write down the domain and the range of the function.

Domain of $f$:

Range of $f$:

[2]

(b) Complete the following table of values (if there exist)

$f(-1)=$	$f(1)=$	$f(2)=$	$f(4)=$
$f(5)=$	$f(6)=$	$f(7)=$	$f(8)=$

[4]

(c) Complete the following table of limits (if there exist)

$\displaystyle \lim_{x\to1^-}f(x)=$	$\displaystyle \lim_{x\to1^+}f(x)=$	$\displaystyle \lim_{x\to1}f(x)=$
$\displaystyle \lim_{x\to4^-}f(x)=$	$\displaystyle \lim_{x\to4^+}f(x)=$	$\displaystyle \lim_{x\to4}f(x)=$
$\displaystyle \lim_{x\to6^-}f(x)=$	$\displaystyle \lim_{x\to6^+}f(x)=$	$\displaystyle \lim_{x\to6}f(x)=$

[9]

Let

$$f(x)=\begin{cases}x^2+1&-1\leq x<1\\4&x=1\\2&1<x\leq4\\x-2&4<x<6\\2&6\leq x<8\end{cases}$$

(a) Write down the domain of the function.

Domain of $f$:

[2]

(b) Complete the following table of values (if there exist)

$f(-1)=$	$f(1)=$	$f(2)=$	$f(4)=$
$f(5)=$	$f(6)=$	$f(7)=$	$f(8)=$

[4]

(c) Complete the following table of limits (if there exist)

$\displaystyle \lim_{x\to1^-}f(x)=\lim_{x\to1^-}(x^2+1)=2$

$\displaystyle \lim_{x\to1^+}f(x)=$

$\displaystyle \lim_{x\to1}f(x)=$

$\displaystyle \lim_{x\to4^-}f(x)=$

$\displaystyle \lim_{x\to4^+}f(x)=$

$\displaystyle \lim_{x\to4}f(x)=$

$\displaystyle \lim_{x\to6^-}f(x)=$

$\displaystyle \lim_{x\to6^+}f(x)=$

$\displaystyle \lim_{x\to6}f(x)=$

[8]