Emma

June 26, 2019

Chapters

Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Solution of exercise 1
Solution of exercise 2
Solution of exercise 3
Solution of exercise 4
Solution of exercise 5
Solution of exercise 6
Solution of exercise 7
Solution of exercise 8

Exercise 1

Find the point in the function y = |x + 2| where it has no derivative. Justify the result by representing it graphically.

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Exercise 2

Find the point in the function y = |x ² − 5x + 6| where it has no derivative. Justify the result by representing it graphically.

Exercise 3

Study the continuity and differentiability of the function defined by:

f(x) = $\begin{cases}cos x + 2 \hspace{1cm} if \hspace{1cm} x < 0 \\ \frac {2x} {\pi} + 1\hspace{1cm} if \hspace{1cm} 0<x<\frac{\pi}{2} \\ sin x + 1 \hspace{1cm} if\hspace{1cm} x \geq \frac {\pi}{2}\end{cases}$

Exercise 4

Given the function:

$f(x) = \begin{cases}3 - ax^2\hspace{1cm} if \hspace{1cm} x \leq 1 \\ \frac {2} {ax} \hspace{1cm} if \hspace{1cm} x>1 \end{cases}$

For what values of a is the function differentiable?

Exercise 5

Determine the values of a and b where the following function is continuous and differentiable:

f(x) = $\begin{cases}x^3 - x\hspace{1cm} if \hspace{1cm} x <0 \\ ax+b \hspace{1cm} if \hspace{1cm} x\geq0\end{cases}$

Exercise 6

Determine the values of a and b for which the function is differentiable at all points:

$f(x) = \begin{cases}bx^2+ax\hspace{1cm} if \hspace{1cm} x \leq -1 \\ \frac {a} {x} \hspace{1cm} if \hspace{1cm} -1<x\leq1 \\ \frac{x^2 + ax + 1} {x + 1} \hspace{1cm} if \hspace{1cm} x > 1\end{cases}$

Exercise 7

Find the points where y = 250 − |x² −1| has no derivative.

Exercise 8

Determine for which values of a and b the function is continuous and differentiable:

$f(x) = \begin{cases}x^2+2\hspace{1cm} if \hspace{1cm} x \leq 0 \\ \sqrt{ax + b} \hspace{1cm} if \hspace{1cm} 0<x\leq2 \\ \frac{-x} {2\sqrt{2}} + \frac {3}{\sqrt{2}} \hspace{1cm} if \hspace{1cm} x > 2\end{cases}$

Solution of exercise 1

Find the point in the function y = |x + 2| where it has no derivative. Justify the result by representing it graphically.

$f(x) = \begin{cases}-x-2\hspace{1cm} if \hspace{1cm} x <-2 \\ x+2 \hspace{1cm} if \hspace{1cm} x\geq-2\end{cases}$

$f(-2)= \lim _ {x \rightarrow -2^{-}} f(x) = \lim _ {x \rightarrow -2^{+}} f(x)=0$

The function is continuous.

$f'(x) = \begin{cases}-1\hspace{1cm} if \hspace{1cm} x <-2 \\ 1 \hspace{1cm} if \hspace{1cm} x>-2\end{cases}$

$f'(-2) ^{-} = -1$ $f'(-2)^{+} = 1$

It has no derivative at P(−2,0).

Solution of exercise 2

Find the point in the function y = |x ² − 5x + 6| where it has no derivative. Justify the result by representing it graphically.

$f(x) = \begin{cases}x^2-5x+6\hspace{1cm} if \hspace{1cm} x <2 \\ -x^2+5x-6 \hspace{1cm} if \hspace{1cm} 2\leq x< 3 \\ x^2-5x+6 \hspace{1cm} if \hspace{1cm} x \geq 3\end{cases}$

$f(2)= \lim _ {x \rightarrow 2^{-}} f(x) = \lim _ {x \rightarrow 2^{+}}f(x) = 0$

$f(3)= \lim _ {x \rightarrow 3^{-}} f(x) = \lim _ {x \rightarrow 3^{+}}f(x) = 0$

The function is continuous.

$f'(x) = \begin{cases}2x - 5\hspace{1cm} if \hspace{1cm} x <2 \\ -2x + 5 \hspace{1cm} if \hspace{1cm} 2<x<3 \\ 2x - 5 \hspace{1cm} if \hspace{1cm} x > 3\end{cases}$

$f'(2) ^ {-} = -1$ $f'(2) ^ {+} = 1$

$f'(3) ^ {-} = -1$ $f' (3) ^ {+} = 1$

The function is not differentiable at: x = 2 and x = 3 or at points P₁(2,0) and P₂(3,0).

Solution of exercise 3

Study the continuity and differentiability of the function defined by:

$f(x) = \begin{cases}cos x + 2\hspace{1cm} if \hspace{1cm} x <0 \\ \frac{2x} {\pi} + 1 \hspace{1cm} if \hspace{1cm} 0<x<\frac{\pi}{2} \\ sin x + 1 \hspace{1cm} if \hspace{1cm} x \geq \frac{\pi}{2}\end{cases}$

The function is not continuous at x = 0 because it has no image. Therefore it is not differentiable.

$f(\frac{\pi}{2}) = sin \frac {\pi}{2} + 1 = 2$

$\lim _ {x \rightarrow (\frac{\pi}{2}) ^ {-}}\frac{2x}{\pi} + 1= \lim _ {x \rightarrow (\frac{\pi}{2}) ^ {+}} sin \frac {\pi}{2} + 1 = 2$

$f(\frac{\pi}{2}) = \lim _ {x \rightarrow (\frac{\pi}{2})}f(x)$

The function is continuous.

$f'(x) = \begin{cases}-sin x\hspace{1cm} if \hspace{1cm} x <0 \\ \frac{2}{\pi} \hspace{1cm} if \hspace{1cm} 0<x<\frac{\pi}{2} \\ cos x \hspace{1cm} if \hspace{1cm} x > \frac{\pi}{2}\end{cases}$

$f'(\frac{\pi}{2}) ^ {-}= \frac{2} {\pi}$ $f'(\frac{\pi}{2}) ^ {+} = cos \frac {\pi}{2} = 0$

The function is not differentiable at any point.

Solution of exercise 4

Given the function:

$f(x) = \begin{cases}3 - ax ^2\hspace{1cm} if \hspace{1cm} x \leq1 \\ \frac{2}{ax} \hspace{1cm} if \hspace{1cm} x>1\end{cases}$

For what values of a is the function differentiable?

$f(1) = 3 - a$

$\lim _ {x \rightarrow 1 ^ {-}}(3 - ax^2) = 3 - a$

$\lim _ {x \rightarrow 1 ^ {+}} \frac {2}{ax} =\frac {2}{a}$

$3 - a = \frac {2}{a}$ $a ^2 - 3a - 2 = 0$

$a = 1$ $a = 2$

$f'(x) = \begin{cases}-2ax\hspace{1cm} if \hspace{1cm} x <1 \\ \frac{-2}{ax^2} \hspace{1cm} if \hspace{1cm} x>1\end{cases}$

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

$-2a = \frac {-2}{a}$ $a ^2 = 1$ $a \pm 1$

Differentiable at a = 1

For x = −1, it is not continuous.

Solution of exercise 5

Determine the values of a and b where the following function is continuous and differentiable:

$f(x) = \begin{cases}x^3 - x\hspace{1cm} if \hspace{1cm} x <0 \\ ax+b \hspace{1cm} if \hspace{1cm} x\geq 0\end{cases}$

$f(0) = b$

$\lim _ {x \rightarrow 0 ^ {-}}(x^3 - x) = 0$

$\lim _ {x \rightarrow 0 ^ {+}} (ax + b) = b$

$b = 0$

$f'(x) = \begin{cases}3x^2 - 1\hspace{1cm} if \hspace{1cm} x <0 \\ a\hspace{1cm} if \hspace{1cm} x>0\end{cases}$

$f'(0^{-}) = -1 f'(0 ^ {+}) = a$

$a = -1$

Solution of exercise 6

Determine the values of a and b for which the function is differentiable at all points:

$f(x) = \begin{cases}bx^2 + ax\hspace{1cm} if \hspace{1cm} x \leq - 1 \\ \frac{a}{x}\hspace{1cm} if \hspace{1cm} -1 < x \leq 1 \\ \frac {x ^2 + ax + 1} {x + 1} \hspace{1cm} if \hspace{1cm} x >1\end{cases}$

A differentiable function has to be continuous. In this case the function is not continuous for x = 0, that is to say, there are no values for a and b which make the function continuous.

Therefore, there are no values of a and b for which the function is differentiable.

Solution of exercise 7

Find the points where y = 250 − |x² −1| has no derivative.

$x ^2 - 1= 0$ $x = \pm 1$

$x \hspace{1cm} (\infty, 1)\hspace{1cm} \hspace{1cm}(-1, 1) \hspace{1cm} (1, \infty)$

$f(x)\hspace\hspace{1cm} + \hspace\hspace{1cm} - \hspace\hspace{1cm} +$

$f(x) = \begin{cases}250 - (x^2 - 1)\hspace{1cm} if \hspace{1cm} x < - 1 \\ 250 + (x ^2 - 1)\hspace{1cm} if \hspace{1cm} -1 \leq x \leq 1 \\ 250 - (x ^2 - 1)\hspace{1cm} if \hspace{1cm} x\geq1\end{cases}$

$f(-1) = 250 + (1^2 - 1) = 250$

$\lim_{x\rightarrow 1 ^ {-}} 250 - (x ^2 - 1) = 250$

$\lim_{x \rightarrow 1 ^ {+}} 250 + (x ^2 - 1) = 250$

$f(1) = 250 - (1^2 - 1) = 250$

$\lim_{x \rightarrow 1 ^ {-}} 250 + (x^2 - 1) = 250$

$\lim_{x \rightarrow 1 ^ {+}} 250 - (x ^2 - 1) = 250$

The function is continuous.

$f'(x) = \begin{cases}-2x\hspace{1cm} if \hspace{1cm} x < - 1 \\ 2x\hspace{1cm} if \hspace{1cm} -1 \leq x < 1 \\ -2x\hspace{1cm} if \hspace{1cm} x\geq1\end{cases}$

$f'(-1^{-}) = 2$ $f'(-1 ^ {+}) = -2$

$f'(1^{-}) = 2$ $f'(1 ^ {+}) = -2$

Is not differentiable at x = −1 and x = 1.

Solution of exercise 8

Determine for which values of a and b the function is continuous and differentiable:

$f(x) = \begin{cases}x^2 + 2\hspace{1cm} if \hspace{1cm} x \leq 0 \\ \sqrt{ax + b}\hspace{1cm} if \hspace{1cm} 0<x\leq 2 \\ \frac{-x}{2\sqrt{2}} + \frac {3} {\sqrt{2}}\hspace{1cm} if \hspace{1cm} x>2\end{cases}$

$f(0) = 2$

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

$\lim_{x \rightarrow 0 ^{+}} \sqrt{ax + b} = \sqrt{b}$

$\sqrt{b} = 2$ $b = 4$

$f(2) = \sqrt{2a + 4}$

$\lim_{x \rightarrow 2 ^ {-}} \sqrt{2a + b} = \sqrt {2a + 4}$

$\lim_{x \rightarrow 2 ^ {+}} \frac {-2} {2 \sqrt{2}} + \frac {3} {\sqrt{2}} = \sqrt{2}$

$\sqrt{2a + 4} = \sqrt{2}$ $a = -1$

For a = −1 and b = 4, the function is continuous.

$f(x) = \begin{cases}2x\hspace{1cm} if \hspace{1cm} x < 0 \\ \frac{1} {2\sqrt{-x + 4}}\hspace{1cm} if \hspace{1cm} 0<x< 2 \\ \frac{-1}{2\sqrt{2}} \hspace{1cm} if \hspace{1cm} x>2\end{cases}$

$f'(0^{-}) = 0$ $f'(0 ^ {+}) = -\frac {1}{4}$

It is not differentiable at x = 0.

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

It is differentiable at x = 2.

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

Derivative Problems

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Solution of exercise 1

Solution of exercise 2

Solution of exercise 3

Solution of exercise 4

Solution of exercise 5

Solution of exercise 6

Solution of exercise 7

Solution of exercise 8

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