# Francesco Giannino » 9.Higher Order Derivatives. Applications of differentiation: local and absolute extremes of a function

### Higher Order Derivatives

Derivatives of order two or more are called higher derivatives and are represented by the following notation:

$f''(x)=\frac{d^{2}f}{dx^{2}},\,f'''(x)=\frac{d^{3}f}{dx^{3}},\,...,f^{(n)}(x)=\frac{d^{n}f}{dx^{n}}$

The last one is read as “the nth derivative of f with respect to x.”
The definition is given as follows by induction:

$f''(x)=\frac{d^{2}f}{dx^{2}}=\frac{d}{dx}(\frac{df}{dx}),\,...,f^{(n)}(x)=\frac{d^{n}f}{dx^{n}}=\frac{d}{dx}(\frac{df^{n-1}}{dx^{n-1}})$

### Exercises

Exercise 1. Compute the second derivative for the following function:

$f(x)=\cos (4x^{2})$

Solution. First we compute the first derivative and then the second derivative for the assigned function:

$f'(x)=-\sin (4x^{2}) \cdot 8x = -8x\sin (4x^{2})$

$f''(x)=-8\sin (4x^{2})-8x\cos(4x^{2})\cdot 8x$

$\Rightarrow f''(x)=-8\sin (4x^{2})-64x^{2}\cos(4x^{2})=$

$=-8[\sin (4x^{2})+8x^{2}\cos(4x^{2})]$

### Exercises

Exercise 2. Compute the second derivative for the following function:

$f(x)=\ln(x^{2})$

Solution. We compute the first derivative and then the second derivative for the assigned function:

$f'(x)=\frac{1}{x^{2}}\cdot 2x=\frac{2}{x}$

$\Rightarrow f''(x)=-\frac{2}{x^{2}}$

### Exercises

Exercise 3. Compute the second derivative for the following function:

$f(x)=\exp(x^{2})$

Solution. We compute the first derivative and then the second derivative for the assigned function:

$f'(x)=\exp(x^{2})\cdot 2x=2x\exp(x^{2})$

$\Rightarrow f''(x)=2\exp(x^{2}) + 2x\exp(x^{2})\cdot 2x=$

$=2\exp(x^{2}) + 4x^{2}\exp(x^{2})$

$\Rightarrow f''(x)=2\exp(x^{2})(1+2x^{2}).$

### Exercises

Exercise 4. Compute the second derivative for the following function:

$f(x)=\log_{10}(x^{2}+1)$

Solution. We compute before the first derivative and then the second derivative for the assigned function:

$f'(x)=\frac{1}{\ln10}\frac{1}{x^{2}+1}\cdot 2x =\frac{2}{\ln10}\cdot\frac{x}{x^{2}+1}$

$\Rightarrow f''(x)=\frac{2}{\ln10}\cdot\frac{x^{2}+1-x(2x)}{(x^{2}+1)^{2}}=$

$=\frac{2}{\ln10}\cdot\frac{x^{2}+1-2x^{2}}{(x^{2}+1)^{2}}$

$\Rightarrow f''(x)=\frac{2}{\ln10}\cdot\frac{1-x^{2}}{(x^{2}+1)^{2}}.$

### Applications of differentiation: local and absolute extremes of a function

Definition. A function f with domain D is said to have an absolute maximum at c if f(x) ≤ f(c) for all x $\in$ D. The number f(c) is called the absolute maximum of f on D. The function f is said to have a local maximum (or relative maximum) at c if there is some open interval (a, b) containing c and f(c) is the absolute maximum of f on (a, b).
Definition. A function f with domain D is said to have an absolute minimum at c if f(x) ≥ f(c) for all x $\in$ D. The number f(c) is called the absolute minimum of f on D. The function f is said to have a local minimum (or relative minimum) at c if there is some open interval (a, b) containing c and f(c) is the absolute minimum of f on (a, b).

### Applications of differentiation: local and absolute extremes of a function

Definition. An absolute maximum or absolute minimum of f is called an absolute extremum of f. A local maximum or minimum of f is called a local extremum of f.
Theorem 1. (Extreme Value Theorem). If a function f is continuous on a closed and bounded interval [a,b], then there exist two points, c1 and c2, in [a,b] such that f(c1) is the absolute minimum of f on [a,b] and f(c2) is the absolute maximum of f on [a,b].

Theorem 2. If f is defined on an open interval (a,b) containing c, f(c) is a local extremum of f and f’(c) exists, then f’(c) = 0.

### Applications of differentiation: local and absolute extremes of a function

Definition. If f is differentiable at c and f’(c) = 0, then we call c a critical point or stationary point of f.

Definition. A function f is said to be increasing on an open interval (a,b) if f(x1)< f(x2) for all x1 and x2 in (a,b) such that x1< x2. The function f is said to be decreasing on (a,b) if f(x1) > f(x2) for all x1 and x2 in (a,b) such that x1< x2. The function f is said to be non-decreasing on (a,b) if f(x1) ≤ f(x2) for all x1 and x2 in (a,b) such that x1< x2. The function f is said to be non-increasing on (a,b) if f(x1)≥ f(x2) for all x1 and x2 in (a,b) such that x1< x2.

### Applications of differentiation: local and absolute extremes of a function

Theorem 3. Suppose that two functions f and g are continuous on a closed and bounded interval [a, b] and are differentiable on the open interval (a,b). Then the following statements are true:

(i) If f’(x) > 0 for each x in (a,b), then f is increasing on (a,b).
(ii) If f’(x) < 0 for each x in (a,b), then f is decreasing on (a, b).
(iii) If f’(x) ≥ 0 for each x in (a b), then f is non-decreasing on (a,b).
(iv) If f’(x) ≤ 0 for each x in (a,b), then f is non-increasing on (a,b).
(v) If f’(x) = 0 for each x in (a, b), then f is constant on (a, b).

### Applications of differentiation: local and absolute extremes of a function

Theorem 4. (First Derivative Test for Extremum). Let f be continuous on an open interval (a, b) and a < c < b.

(i) If f’(x) > 0 on (a,c) and f’(x) < 0 on (c,b), then f(c) is a local maximum of f on (a,b).
(ii) If f’(x) < 0 on (a,c) and f’(x) > 0 on (c,b), then f(c) is a local minimum of f on (a,b).

### Exercises

Exercise 5. Take the function $f(x)=x^{4}-4x^{2},\,\,-3\leq x \leq 3$

and locate the intervals where the graph of f is increasing or decreasing, the local extrema and absolute extrema on [-3, 3].

Solution. The function f(x) is defined for all real numbers. Hence, its domain is: $\forall$x$\in$ [-3, 3]. Now we compute the first derivative to find stationary points and then apply the first derivative test for extrema:

$f'(x)=4x^{3}-8x=0 \Rightarrow 4x(x^{2}-2)=0$

$\Rightarrow x=0 \,\,\,\text{and} x=\pm \sqrt{2}\,\,\,\text{are the stationary points of} \,\,\,f$

### Exercises

To apply the first derivative test for extrema, we solve the following inequality:

$f'(x)=4x^{3}-8x\geq 0 \Rightarrow 4x(x^{2}-2)\geq 0$

$\Rightarrow x\geq 0 \,\,\text{and}\,\,x\leq -\sqrt{2}\cup x \geq \sqrt{2}$

and evaluate the sign of f’.

The function f is decreasing on $] - \infty, - \sqrt 2 [\cup]0, \sqrt 2 [$ and is increasing on $]-2,0 [\cup]\sqrt 2, + \infty [$

### Exercises

Then, applying the first derivative test for extrema, we say:

$f(0)=0 \,\,\text{is a local maximum}$

$f(-\sqrt{2})=f(\sqrt{2})=-8 \,\,\text{is a local minimum}$

Now we evaluate the function f at the end points of the interval [-3, 3]:

$f(-3)=f(3)=45 > 0 = f(0) \,\,\text{is the absolute maximum of f}$

$\text{and it is obtained at the end points of the interval}\,\,[-3,3]$

$\text{while}\,\,f(-\sqrt{2})=f(\sqrt{2})=-8 \,\,\text{is the absolute minimum of f}$

$\text{on the interval}\,\,[-3,3].$

### Exercises

Exercise 6. Take the function $f(x)=\frac{x}{1+x^{2}}$ and locate the intervals where the graph of f is increasing or decreasing, the local extrema and absolute extrema on function’s domain.

Solution. The function f(x) is defined for all real number. Hence, its domain is: $\forall$x$\in$ ] – ∞, + ∞ [. Now we compute the first derivative to find stationary points and then apply the first derivative test for extrema:

$f'(x)=\frac{(1+x^{2})-x(2x)}{(1+x^{2})^{2}}= \frac{1+x^{2}-2x^{2}}{(1+x^{2})^{2}}\Rightarrow$

$\Rightarrow f'(x)=\frac{1-x^{2}}{(1+x^{2})^{2}}$

### Exercises

So f’(x) = 0 when 1 – x2 = 0. Thus the function f has two critical points: $x=-1\,\,\text{and}\,\,x=1$

Now, to apply the first derivative test for extrema, we solve the following inequality:

$f'(x)=\frac{1-x^{2}}{(1+x^{2})^{2}}\geq 0 \Rightarrow 1-x^{2}\geq 0 \Rightarrow
x^{2}-1 \leq 0$

$\Rightarrow -1\leq x \leq 1$

and evaluate the sign of f’.

### Exercises

$\text{Then the function}\,\,f\,\,\text{is decreasing on}\,\,]-\infty,-1[\cup ]1,+\infty[\,\,$

$\text{and is increasing on}\,\,]-1,+1[$

Now, applying the first derivative test for extrema, we say:

$f(-1)=-\frac{1}{2} \,\,\text{is a local minimum at}\,\, x = -1$

$f(1)=\frac{1}{2} \,\,\text{is a local maximum at}\,\, x = 1$

### Exercises

Moreover, we study the behavior of the function f at the end points of its domain ] -, + [:

$\lim_{x\rightarrow - \infty}f(x)=\lim_{x\rightarrow - \infty}\frac{x}{1+x^{2}}=$

$=\lim_{x\rightarrow - \infty}\frac{x}{x^{2}(\frac{1}{x^{2}}+1)}=\lim_{x\rightarrow - \infty}\frac{1}{x(\frac{1}{x^{2}}+1)}= 0$

$\lim_{x\rightarrow +\infty}f(x)=\lim_{x\rightarrow +\infty}\frac{x}{1+x^{2}}=$

$=\lim_{x\rightarrow +\infty}\frac{x}{x^{2}(\frac{1}{x^{2}}+1)}=\lim_{x\rightarrow +\infty}\frac{1}{x(\frac{1}{x^{2}}+1)}= 0$

Showing that the x-axis is a horizontal asymptote for the graph of f.

### Exercises

Now, putting these observations together, we can say that:

$f(-1)=-\frac{1}{2} \,\,\text{is an absolute minimum at}\,\, x = -1$

$f(1)=\frac{1}{2} \,\,\text{is an absolute maximum at}\,\, x = 1.$

### Exercises

Exercise 7. Take the function $f(x)=5x^{3}-3x^{5}$ and locate the intervals where the graph of f is increasing or decreasing, the local extrema and absolute extrema on function’s domain.

Progetto "Campus Virtuale" dell'Università degli Studi di Napoli Federico II, realizzato con il cofinanziamento dell'Unione europea. Asse V - Società dell'informazione - Obiettivo Operativo 5.1 e-Government ed e-Inclusion

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