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.