110 marksNumericalPrecise definition of LimitAnswer

A function is defined by

$f(x) = \begin{cases} x+2, & x<0 \\ 1-x, & x>0 \end{cases}$

Calculate $f(-1)$ , $f(3)$ , and sketch the graph. Prove that the limit does not exist.

$\lim_{x \to 0} \frac{|x|}{x}$

[5+0+5]

A function is defined by

$$f(x) = \begin{cases} x+2, & x<0 \ 1-x, & x>0 \end{cases}$$

Calculate $f(-1)$, $f(3)$, and sketch the graph. Prove that the limit does not exist.

$$\lim_{x \to 0} \frac{|x|}{x}$$

[5+0+5]

Given data: Piecewise function: $f(x) = \begin{cases} x+2, & x<0 \\ 1-x, & x0 \end{cases}$ Note: $f(0)$ is undefined. Tasks: - Compute $f(-1)$ and $f(3)$ , sketch graph [5 marks] - Prove $\displaystyle\lim{x\to 0}\frac{x}{x}$ does not e...

Given data: Piecewise function: $$f(x) = \begin{cases} x+2, & x<0 \ 1-x, & x0 \end{cases}$$ Note: $f(0)$ is undefined. Tasks: - Compute $f(-1)$ and $f(3)$, sketch graph [5 marks] - Prove $\displaystyle\lim{x\to 0}\frac{x}{x}$ does not e...

210 marksNumericalReview of derivativeAnswer

Find the derivative of $f(x) = \sqrt{x}$ . State the domain of $f$ . Estimate the area between the curve and the line $x = 0$ and $x = 2$ where curve is $y^2 = x$ . [3+2+5]

Find the derivative of $f(x) = \sqrt{x}$. State the domain of $f$. Estimate the area between the curve and the line $x = 0$ and $x = 2$ where curve is $y^2 = x$. [3+2+5]

Function: $f(x) = \sqrt{x}$ - Curve: $y^2 = x$ - Bounds: $x = 0$ and $x = 2$ - Marks split: [3 + 2 + 5] --- (a) Derivative of $f(x) = \sqrt{x}$ Using first principles: $f'(x) = \lim{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$ Multiply ...

Function: $f(x) = \sqrt{x}$ - Curve: $y^2 = x$ - Bounds: $x = 0$ and $x = 2$ - Marks split: [3 + 2 + 5] --- (a) Derivative of $f(x) = \sqrt{x}$ Using first principles: $$f'(x) = \lim{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$ Multiply ...

310 marksNumericalConvergence tests and power seriesAnswer

Exam Questions

Question 1: For what values of x does the series converge? $\sum_{n=1}^{\infty} \frac{(x-3)^n}{x}$

Question 2: Calculate $\iint_{R} f(x, y) dA$ , for $f(x,y) = 100 - 6x^2y$ , and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$ . [5+5]

Exam Questions

Question 1: For what values of x does the series converge? $$\sum_{n=1}^{\infty} \frac{(x-3)^n}{x}$$

Question 2: Calculate $\iint_{R} f(x, y) dA$, for $f(x,y) = 100 - 6x^2y$, and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$. [5+5]

(a) Convergence of $\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ - Series: $\displaystyle\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ Since $x$ is independent of the summation index $n$ , factor out $\frac{1}{x}$ : $$\sum{n=1}^{\infty} \frac{(x-3)^n}{x}...

(a) Convergence of $\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ - Series: $\displaystyle\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ Since $x$ is independent of the summation index $n$, factor out $\frac{1}{x}$: $$\sum{n=1}^{\infty} \frac{(x-3)^n}{x}...

410 marksNumericalTaylor's and Maclaurin's seriesAnswer

Find the Maclaurin series for $e^x$ and prove that it represents $e^x$ for all x. Define initial value problem. Solve that initial value problem of $y' + 5y = 1$ , $y(0) = 2$ . Find the volume of a sphere of radius r. [4+4+2]

Find the Maclaurin series for $e^x$ and prove that it represents $e^x$ for all x. Define initial value problem. Solve that initial value problem of $y' + 5y = 1$, $y(0) = 2$. Find the volume of a sphere of radius r. [4+4+2]

(a) Maclaurin Series for $e^x$ and Proof The Maclaurin series of a function is: $f(x) = \sum{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n$ For $f(x) = e^x$ , all derivatives equal $e^x$ : $$f^{(n)}(x) = e^x \implies f^{(n)}(0) = 1 \quad \tex...

(a) Maclaurin Series for $e^x$ and Proof The Maclaurin series of a function is: $$f(x) = \sum{n=0}^{\infty} \frac{f^{(n)}(0)}{n!},x^n$$ For $f(x) = e^x$, all derivatives equal $e^x$: $$f^{(n)}(x) = e^x \implies f^{(n)}(0) = 1 \quad \tex...

55 marksNumericalCombination of functionsAnswer

If $f(x) = \sqrt{x}$ and $g(x) = \sqrt{3-x}$ , find $g \circ f$ and $f \circ g$ . [5]

If $f(x) = \sqrt{x}$ and $g(x) = \sqrt{3-x}$, find $g \circ f$ and $f \circ g$. [5]

$f(x) = \sqrt{x}$ - $g(x) = \sqrt{3-x}$ Required: $g\circ f$ and $f\circ g$ (with domains). All data present and readable. --- $(g\circ f)(x) = g(f(x)) = g(\sqrt{x}) = \sqrt{3 - \sqrt{x}}$ Domain: For $f(x)=\sqrt{x}$ : need $x \geq 0$ ...

$f(x) = \sqrt{x}$ - $g(x) = \sqrt{3-x}$ Required: $g\circ f$ and $f\circ g$ (with domains). All data present and readable. --- $$(g\circ f)(x) = g(f(x)) = g(\sqrt{x}) = \sqrt{3 - \sqrt{x}}$$ Domain: For $f(x)=\sqrt{x}$: need $x \geq 0$...

65 marksNumericalContinuityAnswer

Use continuity to evaluate the limit, $\lim_{x \to 4} \frac{5 + \sqrt{x}}{\sqrt{5 + x}}$ [5]

$\lim{x \to 4} \frac{5 + \sqrt{x}}{\sqrt{5 + x}}$ Point of evaluation: $a = 4$ --- Root functions, polynomial functions, and their sums, products, and quotients are continuous on their domains. If $f$ is continuous at $x = a$ , then: $$...

$$\lim{x \to 4} \frac{5 + \sqrt{x}}{\sqrt{5 + x}}$$ Point of evaluation: $a = 4$ --- Root functions, polynomial functions, and their sums, products, and quotients are continuous on their domains. If $f$ is continuous at $x = a$, then: $$...

75 marksNumericalMean value theoremAnswer

Verify Mean value theorem of $f(x) = x^3 - 3x + 3$ for [-1,2]. [5]

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 3$ on $[-1, 2]$

Given Data

Function: $f(x) = x^3 - 3x + 3$
Interval: $[a, b] = [-1, 2]$ , so $a = -1$ , $b = 2$

Statement of MVT

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then there exists $c \in (a,b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$

Step 1: Verify Conditions

$f(x)$ is a polynomial, hence:

Continuous on $[-1, 2]$ ✓
Differentiable on $(-1, 2)$ ✓

Both conditions satisfied.

Step 2: Compute $f(a)$ and $f(b)$

$f(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5$ $f(2) = (2)^3 - 3(2) + 3 = 8 - 6 + 3 = 5$

Step 3: Compute the Average Slope

$\frac{f(2) - f(-1)}{2 - (-1)} = \frac{5 - 5}{3} = 0$

Step 4: Solve $f'(c) = 0$

$f'(x) = 3x^2 - 3$ $3c^2 - 3 = 0 \implies c^2 = 1 \implies c = \pm 1$

Step 5: Select $c$ in $(-1, 2)$

$c = -1$ : endpoint, not in open interval ✗
$c = 1$ : lies in $(-1, 2)$ ✓

Conclusion

There exists $c = 1 \in (-1, 2)$ such that $f'(1) = 3(1)^2 - 3 = 0 = \frac{f(2) - f(-1)}{2 - (-1)}$

Hence the Mean Value Theorem is verified with $c = 1$ .

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 3$ on $[-1, 2]$

Given Data

Function: $f(x) = x^3 - 3x + 3$
Interval: $[a, b] = [-1, 2]$, so $a = -1$, $b = 2$

Statement of MVT

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Step 1: Verify Conditions

$f(x)$ is a polynomial, hence:

Continuous on $[-1, 2]$ ✓
Differentiable on $(-1, 2)$ ✓

Both conditions satisfied.

Step 2: Compute $f(a)$ and $f(b)$

$$f(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5$$ $$f(2) = (2)^3 - 3(2) + 3 = 8 - 6 + 3 = 5$$

Step 3: Compute the Average Slope

$$\frac{f(2) - f(-1)}{2 - (-1)} = \frac{5 - 5}{3} = 0$$

Step 4: Solve $f'(c) = 0$

$$f'(x) = 3x^2 - 3$$ $$3c^2 - 3 = 0 \implies c^2 = 1 \implies c = \pm 1$$

Step 5: Select $c$ in $(-1, 2)$

$c = -1$: endpoint, not in open interval ✗
$c = 1$: lies in $(-1, 2)$ ✓

Conclusion

There exists $c = 1 \in (-1, 2)$ such that $$f'(1) = 3(1)^2 - 3 = 0 = \frac{f(2) - f(-1)}{2 - (-1)}$$

Hence the Mean Value Theorem is verified with $c = 1$.

85 marksNumericalCurve sketchingAnswer

Sketch the curve $y = x^3 + x$ . [5]

Sketch the curve $y = x^3 + x$. [5]

As a polynomial, the domain is all real numbers $(-\infty, \infty)$ . Since it is strictly increasing (shown below) and unbounded, the range is also $(-\infty, \infty)$ . x-intercept (set $y=0$ ): $x^3 + x = 0 \implies x(x^2+1) = 0$ Since...

As a polynomial, the domain is all real numbers $(-\infty, \infty)$. Since it is strictly increasing (shown below) and unbounded, the range is also $(-\infty, \infty)$. x-intercept (set $y=0$): $$x^3 + x = 0 \implies x(x^2+1) = 0$$ Since...

95 marksNumericalImproper integralsAnswer

Determine whether the integral is convergent or divergent. $\int_1^{\infty} \frac{1}{x} dx$ [5]

Determine whether the integral is convergent or divergent. $$\int_1^{\infty} \frac{1}{x} dx$$ [5]

Integral: $\displaystyle \int1^{\infty} \frac{1}{x}\, dx$ - Lower limit: $1$ , Upper limit: $\infty$ - Integrand: $\frac{1}{x}$ Since the upper limit is infinite, this is an improper integral of Type 1: $$\int1^{\infty} \frac{1}{x}, dx...

Integral: $\displaystyle \int1^{\infty} \frac{1}{x}, dx$ - Lower limit: $1$, Upper limit: $\infty$ - Integrand: $\frac{1}{x}$ Since the upper limit is infinite, this is an improper integral of Type 1: $$\int1^{\infty} \frac{1}{x}, dx...

105 marksNumericalArc lengthAnswer

Find the length f the arc of the semicubical $y^2 = x^2$ between the points (1,1) and (4,8). [5]

Curve: semicubical parabola. The typed equation reads $y^2 = x^2$ but this is a transcription error. The correct semicubical parabola is $y^2 = x^3$ . - Endpoints: $(1, 1)$ and $(4, 8)$ Consistency check: For $y^2 = x^3$ : at $x=1$ ,

Curve: semicubical parabola. The typed equation reads $y^2 = x^2$ but this is a transcription error. The correct semicubical parabola is $y^2 = x^3$. - Endpoints: $(1, 1)$ and $(4, 8)$ Consistency check: For $y^2 = x^3$: at $x=1$,

115 marksNumericalConvergence tests and power seriesAnswer

Test the convergence of the series $\sum_{n=1}^{\infty} \frac{n^n}{n!}$ [5]

Series to test: $\sum{n=1}^{\infty} un, \qquad un = \frac{n^n}{n!}$ Because the general term involves a factorial $n!$ , D'Alembert's Ratio Test is most convenient. $un = \frac{n^n}{n!}, \qquad u{n+1} = \frac{(n+1)^{n+1}}{(n+1)!}$ For...

Series to test: $$\sum{n=1}^{\infty} un, \qquad un = \frac{n^n}{n!}$$ Because the general term involves a factorial $n!$, D'Alembert's Ratio Test is most convenient. $$un = \frac{n^n}{n!}, \qquad u{n+1} = \frac{(n+1)^{n+1}}{(n+1)!}$$ For...

125 marksNumericalDot product and cross ProductAnswer

Define cross product of two vectors. If $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 2\hat{i} + 7\hat{j} - 5\hat{k}$ , find the vector $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$ . [5]

Define cross product of two vectors. If $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 2\hat{i} + 7\hat{j} - 5\hat{k}$, find the vector $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$. [5]

Cross Product of Two Vectors

Given data

$\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k} \quad\Rightarrow\quad (a_1, a_2, a_3) = (1, 3, 4)$ $\vec{b} = 2\hat{i} + 7\hat{j} - 5\hat{k} \quad\Rightarrow\quad (b_1, b_2, b_3) = (2, 7, -5)$

Definition

The cross product (vector product) of two vectors $\vec{a}$ and $\vec{b}$ is a vector perpendicular to the plane containing both, defined as:

$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \, \hat{n}$

where $\theta$ is the angle between the vectors and $\hat{n}$ is the unit vector perpendicular to both, directed by the right-hand rule. In component (determinant) form:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$

Computing $\vec{a} \times \vec{b}$

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 4 \\ 2 & 7 & -5 \end{vmatrix}$

Expanding along the first row:

$= \hat{i}[(3)(-5) - (4)(7)] - \hat{j}[(1)(-5) - (4)(2)] + \hat{k}[(1)(7) - (3)(2)]$

$= \hat{i}(-15 - 28) - \hat{j}(-5 - 8) + \hat{k}(7 - 6)$

$= -43\hat{i} + 13\hat{j} + \hat{k}$

$\boxed{\vec{a} \times \vec{b} = -43\hat{i} + 13\hat{j} + \hat{k}}$

Computing $\vec{b} \times \vec{a}$

By the anti-commutative property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$ :

$\vec{b} \times \vec{a} = 43\hat{i} - 13\hat{j} - \hat{k}$

Direct check:

$\vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -5 \\ 1 & 3 & 4 \end{vmatrix}$

$= \hat{i}[(7)(4)-(-5)(3)] - \hat{j}[(2)(4)-(-5)(1)] + \hat{k}[(2)(3)-(7)(1)]$

$= \hat{i}(28+15) - \hat{j}(8+5) + \hat{k}(6-7) = 43\hat{i} - 13\hat{j} - \hat{k} \;\checkmark$

$\boxed{\vec{b} \times \vec{a} = 43\hat{i} - 13\hat{j} - \hat{k}}$

Summary

Result	Value
$\vec{a} \times \vec{b}$	$-43\hat{i} + 13\hat{j} + \hat{k}$
$\vec{b} \times \vec{a}$	$43\hat{i} - 13\hat{j} - \hat{k}$

Confirming $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ .

Cross Product of Two Vectors

Given data

$$\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k} \quad\Rightarrow\quad (a_1, a_2, a_3) = (1, 3, 4)$$ $$\vec{b} = 2\hat{i} + 7\hat{j} - 5\hat{k} \quad\Rightarrow\quad (b_1, b_2, b_3) = (2, 7, -5)$$

Definition

The cross product (vector product) of two vectors $\vec{a}$ and $\vec{b}$ is a vector perpendicular to the plane containing both, defined as:

$$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta , \hat{n}$$

where $\theta$ is the angle between the vectors and $\hat{n}$ is the unit vector perpendicular to both, directed by the right-hand rule. In component (determinant) form:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}$$

Computing $\vec{a} \times \vec{b}$

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & 4 \ 2 & 7 & -5 \end{vmatrix}$$

Expanding along the first row:

$$= \hat{i}[(3)(-5) - (4)(7)] - \hat{j}[(1)(-5) - (4)(2)] + \hat{k}[(1)(7) - (3)(2)]$$

$$= \hat{i}(-15 - 28) - \hat{j}(-5 - 8) + \hat{k}(7 - 6)$$

$$= -43\hat{i} + 13\hat{j} + \hat{k}$$

$$\boxed{\vec{a} \times \vec{b} = -43\hat{i} + 13\hat{j} + \hat{k}}$$

Computing $\vec{b} \times \vec{a}$

By the anti-commutative property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:

$$\vec{b} \times \vec{a} = 43\hat{i} - 13\hat{j} - \hat{k}$$

Direct check:

$$\vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 7 & -5 \ 1 & 3 & 4 \end{vmatrix}$$

$$= \hat{i}[(7)(4)-(-5)(3)] - \hat{j}[(2)(4)-(-5)(1)] + \hat{k}[(2)(3)-(7)(1)]$$

$$= \hat{i}(28+15) - \hat{j}(8+5) + \hat{k}(6-7) = 43\hat{i} - 13\hat{j} - \hat{k} ;\checkmark$$

$$\boxed{\vec{b} \times \vec{a} = 43\hat{i} - 13\hat{j} - \hat{k}}$$

Summary

Result	Value
$\vec{a} \times \vec{b}$	$-43\hat{i} + 13\hat{j} + \hat{k}$
$\vec{b} \times \vec{a}$	$43\hat{i} - 13\hat{j} - \hat{k}$

Confirming $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.

135 marksNumericalLimits at infinityAnswer

Define limit of a function. $\lim_{x \to \infty} \left(x - \sqrt{x}\right)$ [5]

Function to evaluate: $\lim{x \to \infty} \left(x - \sqrt{x}\right)$ - Also asked: definition of limit of a function. --- Let $f(x)$ be a function defined on an open interval containing the point $a$ (except possibly at $a$ itself). Th...

Function to evaluate: $\lim{x \to \infty} \left(x - \sqrt{x}\right)$ - Also asked: definition of limit of a function. --- Let $f(x)$ be a function defined on an open interval containing the point $a$ (except possibly at $a$ itself). Th...

145 marksNumericalMaximum and minimum valuesAnswer

Find the extreme values of $f(x,y) = y^2 - x^2$ . [5]

Find the extreme values of $f(x,y) = y^2 - x^2$. [5]

Function: $f(x,y) = y^2 - x^2$ No other numeric data required; this is a standard extremum problem. $fx = -2x, \qquad fy = 2y$ Set to zero: $-2x = 0 \Rightarrow x = 0, \qquad 2y = 0 \Rightarrow y = 0$ Critical point: $(0, 0)$ . $$f{xx...

Function: $f(x,y) = y^2 - x^2$ No other numeric data required; this is a standard extremum problem. $$fx = -2x, \qquad fy = 2y$$ Set to zero: $$-2x = 0 \Rightarrow x = 0, \qquad 2y = 0 \Rightarrow y = 0$$ Critical point: $(0, 0)$. $$f{xx...

155 marksNumericalSecond order linear differential equationsAnswer

Find the solution of $y'' + 6y' + 9 = 0$ , $y(0) = 2$ , $y'(0) = 1$ . [5]

Find the solution of $y'' + 6y' + 9 = 0$, $y(0) = 2$, $y'(0) = 1$. [5]

ODE: $y'' + 6y' + 9 = 0$ - Initial conditions: $y(0) = 2$ , $y'(0) = 1$ Note: As written, the equation contains a constant term " $+9$ " rather than " $+9y$ ". Read literally, $y'' + 6y' = -9$ is a non-homogeneous equation. However, the sta...

ODE: $y'' + 6y' + 9 = 0$ - Initial conditions: $y(0) = 2$, $y'(0) = 1$ Note: As written, the equation contains a constant term "$+9$" rather than "$+9y$". Read literally, $y'' + 6y' = -9$ is a non-homogeneous equation. However, the sta...

Mathematics I 2074 question paper

Mean Value Theorem Verification for f(x)=x3−3x+3f(x) = x^3 - 3x + 3f(x)=x3−3x+3 on [−1,2][-1, 2][−1,2]

Given Data

Statement of MVT

Step 1: Verify Conditions

Step 2: Compute f(a)f(a)f(a) and f(b)f(b)f(b)

Step 3: Compute the Average Slope

Step 4: Solve f′(c)=0f'(c) = 0f′(c)=0

Step 5: Select ccc in (−1,2)(-1, 2)(−1,2)

Conclusion

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 3$ on $[-1, 2]$

Given Data

Statement of MVT

Step 1: Verify Conditions

Step 2: Compute $f(a)$ and $f(b)$

Step 3: Compute the Average Slope

Step 4: Solve $f'(c) = 0$

Step 5: Select $c$ in $(-1, 2)$

Conclusion

Cross Product of Two Vectors

Given data

Definition

Computing a⃗×b⃗\vec{a} \times \vec{b}a×b

Computing b⃗×a⃗\vec{b} \times \vec{a}b×a

Summary

Cross Product of Two Vectors

Given data

Definition

Computing $\vec{a} \times \vec{b}$

Computing $\vec{b} \times \vec{a}$

Summary

Mean Value Theorem Verification for f(x)=x3−3x+3f(x) = x^3 - 3x + 3f(x)=x3−3x+3 on [−1,2][-1, 2][−1,2]

Given Data

Statement of MVT

Step 1: Verify Conditions

Step 2: Compute f(a)f(a)f(a) and f(b)f(b)f(b)

Step 3: Compute the Average Slope

Step 4: Solve f′(c)=0f'(c) = 0f′(c)=0

Step 5: Select ccc in (−1,2)(-1, 2)(−1,2)

Conclusion

Cross Product of Two Vectors

Given data

Definition

Computing a⃗×b⃗\vec{a} \times \vec{b}a×b

Computing b⃗×a⃗\vec{b} \times \vec{a}b×a

Summary

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 3$ on $[-1, 2]$

Step 2: Compute $f(a)$ and $f(b)$

Step 4: Solve $f'(c) = 0$

Step 5: Select $c$ in $(-1, 2)$

Computing $\vec{a} \times \vec{b}$

Computing $\vec{b} \times \vec{a}$

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 3$ on $[-1, 2]$

Step 2: Compute $f(a)$ and $f(b)$

Step 4: Solve $f'(c) = 0$

Step 5: Select $c$ in $(-1, 2)$

Computing $\vec{a} \times \vec{b}$

Computing $\vec{b} \times \vec{a}$