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Calculus is a branch of mathematics that deals with the study of rates of change and accumulation. It is widely used in various fields such as science, finance, engineering, and industry to model and solve problems that involve change and motion. The two main branches of calculus are differential calculus and integral calculus.

Differential calculus involves the study of rates of change and slopes of curves. It is used to find the instantaneous rate of change of a function, which is the rate of change at a particular point. Differential calculus is also used to find maximum and minimum values of functions, and to determine the concavity and inflection points of a curve.

Integral calculus, on the other hand, is used to find the accumulation of quantities and the area under curves. It is used to find the total change over a period of time, or the total amount of a substance produced in a chemical reaction. Integral calculus is also used to find the volume of irregularly shaped objects, and to calculate probabilities in statistics.

Partial differentiation and multiple integration are extensions of calculus to higher dimensions. Partial differentiation is used to find the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is used extensively in physics and engineering to model and solve problems involving multiple variables. Multiple integration is used to find the volume, surface area, and center of mass of objects in higher dimensions. It is also used in probability theory to calculate the probability of events in higher dimensions.

In summary, calculus is a fundamental tool in mathematical modelling, and is used extensively in various fields. Students who study calculus will gain the ability to accurately and efficiently calculate rates of change and accumulation, and will be able to apply these methods to solve real-world problems. The methods covered in this module will be essential for students throughout their degree, and will prepare them for advanced topics in mathematics and other disciplines.

Find the number $b$ for which $x^b=b^x$ has only one solution (at $x=b$ ).

Let’s first consider the case where $b=1$. In this case, we have $x^b=b^x=1^x=1$ for all $x$. Therefore, there is only one solution, namely $x=1$.

Now, let’s consider the case where $b\neq 1$. If we take the natural logarithm of both sides of $x^b=b^x$, we get:

$b \ln (x)=x \ln (b)$

If we define $f(x)=\frac{x\ln(b)}{b}$, then the equation above can be written as $f(x)=\ln(x)$. Note that $f(x)$ is an increasing function for $x>0$ if $b>1$, and a decreasing function for $0<x<1$ if $0<b<1$.

Therefore, for $b>1$, the graph of $f(x)$ intersects the graph of $\ln(x)$ at most twice, once in the interval $(0,1)$ and once in the interval $(1,\infty)$. Since $f(1)=\ln(b)$, the equation $f(x)=\ln(x)$ has a unique solution in $(1,\infty)$ if and only if $\ln(b)>0$, which is equivalent to $b>1$.

For $0<b<1$, the graph of $f(x)$ intersects the graph of $\ln(x)$ at most twice, once in the interval $(0,1)$ and once in the interval $(1,\infty)$. Since $f(1)=\ln(b)$, the equation $f(x)=\ln(x)$ has a unique solution in $(0,1)$ if and only if $\ln(b)<0$, which is equivalent to $0<b<1$.

Therefore, the equation $x^b=b^x$ has only one solution if $b=\boxed{1}$ or $0<b<1$.

Graph $y(x)=e^x-x^e$. Locate its minimum.

To find the minimum of the function $y(x)=e^x-x^e$, we need to find its critical points. The critical points occur where the derivative of the function is zero or undefined.

Taking the derivative of $y(x)$ with respect to $x$, we get:

$$y'(x) = e^x – ex^{e-1}$$

To find the critical points, we set $y'(x) = 0$ and solve for $x$:

$$e^x – ex^{e-1} = 0$$

Dividing both sides by $e^x$ gives:

$$1 – \frac{x}{e} = 0$$

Therefore, the critical point is at $x=e$. To confirm that this is a minimum, we need to check the second derivative of $y(x)$:

$$y”(x) = e^x – e(e-1)x^{e-2}$$

Plugging in $x=e$ gives:

$$y”(e) = e^e – e(e-1)e^{e-2} = e^e – e^e + e = e$$

Since $y”(e) > 0$, the critical point at $x=e$ is a minimum.

Therefore, the function $y(x)=e^x-x^e$ has a minimum at $x=e$, and the minimum value is:

$$y(e) = e^e – e^e = 0$$

So the minimum value of the function is 0, which occurs at $x=e$.

Find all real solutions to $x^4-11 x^3+5 x-2=0$.

We can approach this problem by factoring the given polynomial, if possible. One way to factor it is to use the Rational Root Theorem, which states that if a polynomial with integer coefficients has a rational root $\frac{p}{q}$ (in lowest terms), then $p$ must divide the constant term of the polynomial and $q$ must divide the leading coefficient.

In this case, the constant term is $-2$ and the leading coefficient is $1$, so any rational root must be of the form $\pm 1, \pm 2$. Checking these values, we find that none of them are roots of the polynomial.

Since the polynomial is of degree $4$, it must have exactly four complex roots (counting multiplicity). Let $a$, $b$, $c$, and $d$ be the four roots (not necessarily distinct) of the polynomial. Then we can write the polynomial as:

$$x^4 – 11x^3 + 5x – 2 = (x – a)(x – b)(x – c)(x – d)$$

Expanding the right-hand side, we get:

$$x^4 – (a + b + c + d)x^3 + (ab + ac + ad + bc + bd + cd)x^2 – (abc + abd + acd + bcd)x + abcd$$

Comparing coefficients with the left-hand side of the equation, we have:

\begin{align*} a + b + c + d &= 11 \ ab + ac + ad + bc + bd + cd &= 0 \ abc + abd + acd + bcd &= -5 \ abcd &= 2 \end{align*}

We can use these equations to eliminate variables and simplify the problem. For example, we can use the equation $abcd = 2$ to solve for one of the variables in terms of the other three:

$$d = \frac{2}{abc}$$

Substituting this into the equation $a + b + c + d = 11$ gives:

$$a + b + c + \frac{2}{abc} = 11$$

Multiplying both sides by $abc$ and rearranging, we get:

$$abc^2 + ab^2c + a^2bc + 2 = 11abc$$

This is a cubic equation in $abc$, which can be solved using various methods, such as factoring or the cubic formula. However, it turns out that the equation has no rational roots, which means that none of the roots of the polynomial $x^4 – 11x^3 + 5x – 2$ are rational.

Therefore, the only solutions to the equation are the four complex roots, which we can find using numerical methods such as the Newton-Raphson method or by using a computer algebra system.