Solve and explain in English

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QUESTION 1

a) Define the following terms:

i) Function A function is a relation between a set of inputs (domain) and a set of permissible outputs (codomain), where each input is related to exactly one output.
ii) Injective function An injective function (or one-to-one function) $f: A \to B$ is a function where every element of the codomain is mapped to by at most one element of the domain. In other words, if $f(x_1) = f(x_2)$ , then $x_1 = x_2$ .
iii) Surjective function A surjective function (or onto function) $f: A \to B$ is a function where every element of the codomain is mapped to by at least one element of the domain. In other words, for every $y \in B$ , there exists at least one $x \in A$ such that $f(x) = y$ .

b) State the following theorems:

i) Maximum and Minimum Value Theorem (Extreme Value Theorem) If a function $f$ is continuous on a closed and bounded interval $[a, b]$ , then $f$ attains an absolute maximum value and an absolute minimum value on that interval.
ii) Taylor Theorem If a function $f$ is $n$ times differentiable at a point $a$ , then it can be approximated by a Taylor polynomial of degree $n$ around $a$ . Taylor's Theorem with Peano remainder states: $f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + o((x-a)^n)$
iii) Lagrange Remainder Theorem The Lagrange Remainder Theorem provides a specific form for the remainder term in Taylor's Theorem. If $f$ is a function that is $(n+1)$ times continuously differentiable on an interval containing $a$ and $x$ , then the remainder $R_n(x) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$ can be written as: $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$ .

QUESTION 2

a) Define derivative of a function at a point and hence compute the derivative of a function $f(x) = 3x^3 + 2x - 1$ at $x=2$ .

The derivative of a function $f$ at a point $x_0$ is the limit of the average rate of change of the function as the increment $h$ approaches zero. It represents the slope of the tangent line to the graph of $f$ at $x_0$ .

f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h}

To compute the derivative of $f(x) = 3x^3 + 2x - 1$ at $x=2$ :

Find the general derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}(3x^3 + 2x - 1)$ $f'(x) = 3 \cdot 3x^{3-1} + 2 \cdot 1x^{1-1} - 0$ $f'(x) = 9x^2 + 2$
Evaluate the derivative at $x=2$ .
$f'(2) = 9(2)^2 + 2$ $f'(2) = 9(4) + 2$ $f'(2) = 36 + 2$ $f'(2) = \boxed{38}$
b) Define composite function and hence if $f: x \to 3x+1$ , $g: x \to x-1$ , $h: x \to 2x$ . Find (i) $hgf$ (ii) $fgh$ (iii) $ghf$ and find the value of $hgf(4)$ , $fgh(2)$ and $ghf(-2)$ .

A composite function is a function formed by applying one function to the result of another function. If $f$ and $g$ are two functions, the composition $f \circ g$ (read "f of g") is defined by $(f \circ g)(x) = f(g(x))$ .

Given functions:
$f(x) = 3x+1$
$g(x) = x-1$
$h(x) = 2x$
- i) $hgf(x)$

Calculate $gf(x)$ .

$$
gf(x) = g(f(x)) = g(3x+1)
$$
$$
gf(x) = (3x+1) - 1
$$
$$
gf(x) = 3x
$$

Calculate $hgf(x)$ .

$$
hgf(x) = h(gf(x)) = h(3x)
$$
$$
hgf(x) = 2(3x)
$$
$$
hgf(x) = \boxed{6x}
$$

ii) $fgh(x)$

Calculate $gh(x)$ .

$$
gh(x) = g(h(x)) = g(2x)
$$
$$
gh(x) = (2x) - 1
$$
$$
gh(x) = 2x-1
$$

Calculate $fgh(x)$ .

$$
fgh(x) = f(gh(x)) = f(2x-1)
$$
$$
fgh(x) = 3(2x-1) + 1
$$
$$
fgh(x) = 6x - 3 + 1
$$
$$
fgh(x) = \boxed{6x-2}
$$

iii) $ghf(x)$

Calculate $hf(x)$ .

$$
hf(x) = h(f(x)) = h(3x+1)
$$
$$
hf(x) = 2(3x+1)
$$
$$
hf(x) = 6x+2
$$

Calculate $ghf(x)$ .
$$ ghf(x) = g(hf(x)) = g(6x+2) $$ $$ ghf(x) = (6x+2) - 1 $$ $$ ghf(x) = \boxed{6x+1} $$
- Calculate the values:
  
  $hgf(4)$ Using $hgf(x) = 6x$ : $hgf(4) = 6(4) = \boxed{24}$
  
  $fgh(2)$ Using $fgh(x) = 6x-2$ : $fgh(2) = 6(2) - 2$ $fgh(2) = 12 - 2 = \boxed{10}$
  
  $ghf(-2)$ Using $ghf(x) = 6x+1$ : $ghf(-2) = 6(-2) + 1$ $ghf(-2) = -12 + 1 = \boxed{-11}$
c) Define inverse function and hence if $f: x \to 2x+5$ , $g: x \to x+4$ . Find (i) $f^{-1}$ (ii) $g^{-1}$ (iii) $(fg)^{-1}$ (iv) Show that $g^{-1}f^{-1} = (fg)^{-1}$ .

An inverse function, denoted $f^{-1}$ , is a function that "reverses" the action of another function $f$ . If $f(x)=y$ , then $f^{-1}(y)=x$ . For a function to have an inverse, it must be bijective (both injective and surjective).

Given functions:
$f(x) = 2x+5$
$g(x) = x+4$
- i) $f^{-1}(x)$
Set $y = f(x)$ .
$$ y = 2x+5 $$
Solve for $x$ in terms of $y$ .
$$ y-5 = 2x $$ $$ x = \frac{y-5}{2} $$
Replace $y$ with $x$ to get $f^{-1}(x)$ .
$$ f^{-1}(x) = \boxed{\frac{x-5}{2}} $$
- ii) $g^{-1}(x)$
Set $y = g(x)$ .
$$ y = x+4 $$
Solve for $x$ in terms of $y$ .
$$ x = y-4 $$
Replace $y$ with $x$ to get $g^{-1}(x)$ .
$$ g^{-1}(x) = \boxed{x-4} $$
- iii) $(fg)^{-1}(x)$

Calculate $fg(x)$ .

$$
fg(x) = f(g(x)) = f(x+4)
$$
$$
fg(x) = 2(x+4) + 5
$$
$$
fg(x) = 2x + 8 + 5
$$
$$
fg(x) = 2x+13
$$

Set $y = fg(x)$ .
$$ y = 2x+13 $$
Solve for $x$ in terms of $y$ .
$$ y-13 = 2x $$ $$ x = \frac{y-13}{2} $$
Replace $y$ with $x$ to get $(fg)^{-1}(x)$ .
$$ (fg)^{-1}(x) = \boxed{\frac{x-13}{2}} $$
- iv) Show that $g^{-1}f^{-1} = (fg)^{-1}$

Calculate $g^{-1}f^{-1}(x)$ .

$$
g^{-1}f^{-1}(x) = g^{-1}(f^{-1}(x))
$$
We have $f^{-1}(x) = \frac{x-5}{2}$ and $g^{-1}(x) = x-4$.
$$
g^{-1}f^{-1}(x) = g^{-1}\left(\frac{x-5}{2}\right)
$$
$$
g^{-1}f^{-1}(x) = \left(\frac{x-5}{2}\right) - 4
$$
$$
g^{-1}f^{-1}(x) = \frac{x-5}{2} - \frac{8}{2}
$$
$$
g^{-1}f^{-1}(x) = \frac{x-5-8}{2}
$$
$$
g^{-1}f^{-1}(x) = \frac{x-13}{2}
$$

Compare with $(fg)^{-1}(x)$ .
We found $(fg)^{-1}(x) = \frac{x-13}{2}$. Since $g^{-1}f^{-1}(x) = \frac{x-13}{2}$ and $(fg)^{-1}(x) = \frac{x-13}{2}$, we can conclude that $g^{-1}f^{-1} = (fg)^{-1}$.
d) Define derivative of a function at a point and hence compute the derivative of a function $f(x) = 3x^3 + 2x - 1$ at $x=2$ .
This question is identical to question 2a. The solution is the same.

The derivative of a function $f$ at a point $x_0$ is the limit of the average rate of change of the function as the increment $h$ approaches zero. It represents the slope of the tangent line to the graph of $f$ at $x_0$ .
$f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h}$
To compute the derivative of $f(x) = 3x^3 + 2x - 1$ at $x=2$ :
Find the general derivative of $f(x)$ .
$f'(x) = 9x^2 + 2$
Evaluate the derivative at $x=2$ .
$f'(2) = 9(2)^2 + 2 = 36 + 2 = \boxed{38}$
QUESTION 3

a) Define integration of a function and hence solve the following integrals:

Integration of a function is the process of finding the area under the curve of that function. It is the inverse operation of differentiation. The indefinite integral of a function $f(x)$ is a function $F(x)$ whose derivative is $f(x)$ , i.e., $F'(x) = f(x)$ . The definite integral calculates the net area between the function and the x-axis over a given interval.
- i. $\int 2\sqrt{x}\,dx$
Rewrite $\sqrt{x}$ as $x^{1/2}$ .
$$ \int 2x^{1/2}\,dx $$
Apply the power rule for integration $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ .
$$ 2 \int x^{1/2}\,dx = 2 \cdot \frac{x^{1/2+1}}{1/2+1} + C $$ $$ = 2 \cdot \frac{x^{3/2}}{3/2} + C $$ $$ = 2 \cdot \frac{2}{3} x^{3/2} + C $$ $$ = \boxed{\frac{4}{3} x^{3/2} + C} $$
- ii. $\int_{3}^{8} \frac{dx}{\sqrt{1+x}}$
Rewrite the integral with a negative power.
$$ \int_{3}^{8} (1+x)^{-1/2}\,dx $$
Perform a simple substitution.
Let $u = 1+x$ , then $du = dx$ .
Change the limits of integration: if $x=3$ , $u=1+3=4$ . If $x=8$ , $u=1+8=9$ . $\int_{4}^{9} u^{-1/2}\,du$

Integrate using the power rule.

$$
\left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{4}^{9} = \left[ \frac{u^{1/2}}{1/2} \right]_{4}^{9}
$$
$$
= \left[ 2\sqrt{u} \right]_{4}^{9}
$$

Evaluate at the limits.
$$ 2\sqrt{9} - 2\sqrt{4} $$ $$ = 2(3) - 2(2) $$ $$ = 6 - 4 $$ $$ = \boxed{2} $$
- iii. $\int \frac{dx}{2x^2+8x+20}$

Factor the denominator to complete the square.

$$
2x^2+8x+20 = 2(x^2+4x+10)
$$
$$
= 2(x^2+4x+4+6)
$$
$$
= 2((x+2)^2+6)
$$
The integral becomes:
$$
\int \frac{dx}{2((x+2)^2+6)} = \frac{1}{2} \int \frac{dx}{(x+2)^2+6}
$$

Perform a substitution.
Let $u = x+2$ , then $du = dx$ . $\frac{1}{2} \int \frac{du}{u^2+6}$
Use the integration formula $\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$ .
Here $a^2=6$ , so $a=\sqrt{6}$ . $\frac{1}{2} \cdot \frac{1}{\sqrt{6}} \arctan\left(\frac{u}{\sqrt{6}}\right) + C$
Substitute back $u = x+2$ .
$$ = \boxed{\frac{1}{2\sqrt{6}} \arctan\left(\frac{x+2}{\sqrt{6}}\right) + C} $$
b) State fundamental theorem of calculus and hence find the area bounded by the curves $x=y^2$ and $x=2-y^2$ on the closed interval $[-1, 1]$ .

The Fundamental Theorem of Calculus establishes a connection between the operations of differentiation and integration. It consists of two parts:
- Part 1: If $f$ is continuous on $[a, b]$ , then the function $F(x) = \int_{a}^{x} f(t)\,dt$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , and $F'(x) = f(x)$ for all $x \in (a, b)$ .
- Part 2: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (i.e., $F'(x) = f(x)$ ), then $\int_{a}^{b} f(x)\,dx = F(b) - F(a)$ .
To find the area bounded by the curves $x=y^2$ and $x=2-y^2$ on the closed interval $[-1, 1]$ :
We will integrate with respect to $y$ . The interval $[-1, 1]$ is for $y$ .
Find the points of intersection of the two curves.
$y^2 = 2-y^2$ $2y^2 = 2$ $y^2 = 1$ $y = \pm 1$
The intersection points are $y=-1$ and $y=1$ , which match the given interval boundaries.
Determine which function is "to the right" (has a larger $x$ -value) on the interval $[-1, 1]$ .
Let's pick a test point $y=0$ within the interval:
For $x=y^2$ , $x=0^2=0$ .
For $x=2-y^2$ , $x=2-0^2=2$ .
Since $2 > 0$ , the curve $x=2-y^2$ is to the right of $x=y^2$ .
Set up the integral for the area.
The area $A$ is given by the integral of the right function minus the left function, integrated with respect to $y$ .
$A = \int_{-1}^{1} ((2-y^2) - y^2)\,dy$ $A = \int_{-1}^{1} (2 - 2y^2)\,dy$
Integrate the function.
$A = \left[ 2y - \frac{2y^3}{3} \right]_{-1}^{1}$
Evaluate the integral at the limits.
$A = \left( 2(1) - \frac{2(1)^3}{3} \right) - \left( 2(-1) - \frac{2(-1)^3}{3} \right)$ $A = \left( 2 - \frac{2}{3} \right) - \left( -2 - \frac{2(-1)}{3} \right)$ $A = \left( 2 - \frac{2}{3} \right) - \left( -2 + \frac{2}{3} \right)$ $A = 2 - \frac{2}{3} + 2 - \frac{2}{3}$ $A = 4 - \frac{4}{3}$ $A = \frac{12}{3} - \frac{4}{3}$ $A = \boxed{\frac{8}{3}}$

Solve and explain in English

Step-by-step solution

Find the general derivative of f(x)f(x)f(x).

Evaluate the derivative at x=2x=2x=2.

Calculate gf(x)gf(x)gf(x).

Calculate hgf(x)hgf(x)hgf(x).

Calculate gh(x)gh(x)gh(x).

Calculate fgh(x)fgh(x)fgh(x).

Calculate hf(x)hf(x)hf(x).

Calculate ghf(x)ghf(x)ghf(x).

Set y=f(x)y = f(x)y=f(x).

Solve for xxx in terms of yyy.

Replace yyy with xxx to get f−1(x)f^{-1}(x)f−1(x).

Set y=g(x)y = g(x)y=g(x).

Solve for xxx in terms of yyy.

Replace yyy with xxx to get g−1(x)g^{-1}(x)g−1(x).

Calculate fg(x)fg(x)fg(x).

Set y=fg(x)y = fg(x)y=fg(x).

Solve for xxx in terms of yyy.

Replace yyy with xxx to get (fg)−1(x)(fg)^{-1}(x)(fg)−1(x).

Calculate g−1f−1(x)g^{-1}f^{-1}(x)g−1f−1(x).

Compare with (fg)−1(x)(fg)^{-1}(x)(fg)−1(x).

Find the general derivative of f(x)f(x)f(x).

Evaluate the derivative at x=2x=2x=2.

Rewrite x\sqrt{x}x​ as x1/2x^{1/2}x1/2.

Apply the power rule for integration ∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1​+C.

Rewrite the integral with a negative power.

Perform a simple substitution.

Integrate using the power rule.

Evaluate at the limits.

Factor the denominator to complete the square.

Perform a substitution.

Use the integration formula ∫1a2+x2 dx=1aarctan⁡(xa)+C\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C∫a2+x21​dx=a1​arctan(ax​)+C.

Substitute back u=x+2u = x+2u=x+2.

Find the points of intersection of the two curves.

Determine which function is "to the right" (has a larger xxx-value) on the interval [−1,1][-1, 1][−1,1].

Set up the integral for the area.

Integrate the function.

Evaluate the integral at the limits.

Still stuck on this one?

More English Questions

Find the general derivative of $f(x)$ .

Evaluate the derivative at $x=2$ .

Calculate $gf(x)$ .

Calculate $hgf(x)$ .

Calculate $gh(x)$ .

Calculate $fgh(x)$ .

Calculate $hf(x)$ .

Calculate $ghf(x)$ .

Set $y = f(x)$ .

Solve for $x$ in terms of $y$ .

Replace $y$ with $x$ to get $f^{-1}(x)$ .

Set $y = g(x)$ .

Solve for $x$ in terms of $y$ .

Replace $y$ with $x$ to get $g^{-1}(x)$ .

Calculate $fg(x)$ .

Set $y = fg(x)$ .

Solve for $x$ in terms of $y$ .

Replace $y$ with $x$ to get $(fg)^{-1}(x)$ .

Calculate $g^{-1}f^{-1}(x)$ .

Compare with $(fg)^{-1}(x)$ .

Find the general derivative of $f(x)$ .

Evaluate the derivative at $x=2$ .

Rewrite $\sqrt{x}$ as $x^{1/2}$ .

Apply the power rule for integration $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ .

Use the integration formula $\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$ .

Substitute back $u = x+2$ .

Determine which function is "to the right" (has a larger $x$ -value) on the interval $[-1, 1]$ .