*Notes to a video lecture on http://www.unizor.com*

__Derivatives - Problems 1__

*Problem 1.1*

Consider a sufficiently smooth function

**.**

*f(x)*Is condition

*f*^{ I}(x_{0}) = 0(a) necessary,

(b) sufficient or

(c) necessary and sufficient

for

**to have a local extremum (local maximum or local minimum) at point**

*f(x*_{0})**?**

*x=x*_{0}Prove your answer.

*Answer*

(a) necessary

It is not sufficient because it might be an inflection point, like for

**at**

*f(x)=x*^{3}**.**

*x=0**Problem 1.2*

Consider a sufficiently smooth function

**.**

*f(x)*Is condition

*f*^{ II}(x_{0}) = 0(a) necessary,

(b) sufficient or

(c) necessary and sufficient

for

**to have an inflection point at**

*f(x*_{0})**?**

*x=x*_{0}Prove your answer.

*Answer*

(a) necessary

It is not sufficient because it might be a point of local extremum, like for

**at**

*f(x)=x*^{4}**.**

*x=0**Problem 1.3 - Derivative of the inverse function theorem*

Consider a sufficiently smooth function

**with derivative**

*y=f(x)*

*f*^{ I}(x)Prove that its inverse

**(that is,**

*y=g(x)***) has a derivative**

*f(g(x))=x*

*g*^{ I}(x)

*g*^{ I}(x) = 1 / f^{ I}(g(x))*Example 1*

*y = f(x) = x*^{n}**(inverse)**

*y = g(x) = x*^{1/n}

*f*^{ I}(x) = n·x^{n−1}

*g*

= (1/n)·x^{ I}(x) = (1/n)·x^{(1/n)−1}== (1/n)·x

^{(1−n)/n}

*1 / f*

= (1/n)·1/x

= (1/n)·x^{ I}(g(x)) = 1 / {n·[x^{1/n}]^{n−1}} == (1/n)·1/x

^{(n−1)/n}== (1/n)·x

^{(1−n)/n}= g^{ I}(x)*Example 2*

*y = f(x) = e*^{x}**(inverse)**

*y = g(x) = ln(x)*

*f*^{ I}(x) = e^{x}

*g*^{ I}(x) = 1/x

*1 / f*

= 1 / e^{ I}(g(x)) == 1 / e

^{ ln(x)}= 1/x = g^{ I}(x)*Problem 1.4*

Consider all possible regular square prisms with a given surface area.

Under what condition between the length of a base' side

**and altitude**

*a***the volume of this prism will be minimum or maximum?**

*h**Solution*

Volume

*V=a²h*Surface area

*S=2a²+4ah*Since

**is given,**

*S*

*h=(S−2a²)/4a*Substitute it into an expression for volume:

*V(a) = a²·(S−2a²)/4a =*

= (1/4)·a·(S−2a²)= (1/4)·a·(S−2a²)

So, we have to find extremum(s) of function

*V(a) = (1/4)·a·(S−2a²) =*

= (1/4)·(−2a³+S·a)= (1/4)·(−2a³+S·a)

This is a polynomial function of

**, defined on an interval from**

*a***to a maximum value when the volume is still greater or equal to zero, that is satisfying the condition**

*0***.**

*S−2a² ≥ 0*Here is how this function looks on a graph (we have chosen

**in this case, so we have to consider this function only on an interval**

*S=6***[**):

*0,√3*]As seen from the graph, the extremum of our function within the specified domain is a local maximum.

To find its extremum(s), find the stationary points where derivative equals to zero:

*d*=

**V**/dx

*(1/4)·(−6a²+S)*To make sure, we are dealing with a local maximum, we can take a second derivative, it's equal to

**, and it is negative within a domain of our function, which confirms that a stationary point is a local maximum.**

*−3a*Set the first derivative to

**, getting an equation for variable**

*0***:**

*a*

*−6a²+S = 0*Its only root within the established domain is

**.**

*a = √S/6*Now we can find the corresponding value of altitude

**in terms of surface area**

*h***:**

*S***.**

*h = (S−2a²)/4a =*

= [S−(S/3)]/(4√S/6) = √S/6= [S−(S/3)]/(4√S/6) = √S/6

As we see,

**, which means that the maximum volume is reached when our prism is a cube.**

*h=a*
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