#### (a)

Here we have $textbf{graph of this function and a series of secant lines}$ which we can use to estimate the slope of the tangent when $x=2$.

On the next picture are $textbf{slopes of the secant lines}$ in some points and in point (2, f(2)), and we can conlude that the slope of the tangent line when $x=2$ is:

$textbf{slope}$ = $dfrac{f(x)-f(2)}{x-2}$

#### (b)

Here we have $textbf{graph of this function and a series of secant lines}$ which we can use to estimate the slope of the tangent when $x=2$.

On the next picture are $textbf{slopes of the secant lines}$ in some points and in point (2, f(2)), and we can conlude that the slope of the tangent line when $x=2$ is:

$textbf{slope}$ = $dfrac{f(x)-f(2)}{x-2}$

#### (c)

Here we have $textbf{graph of this function and a series of secant lines}$ which we can use to estimate the slope of the tangent when $x=2$.

On the next picture are $textbf{slopes of the secant lines}$ in some points and in point (2, f(2)), and we can conlude that the slope of the tangent line when $x=2$ is:

$textbf{slope}$ = $dfrac{f(x)-f(2)}{x-2}$

#### (d)

Here we have $textbf{graph of this function and a series of secant lines}$ which we can use to estimate the slope of the tangent when $x=2$.Here, we can see that all secant lines are all equivalent to the given function.

On the next picture are $textbf{slopes of the secant lines}$ in some points and in point (2, f(2)), and we can conlude that the slope of the tangent line when $x=2$ is:

$textbf{slope}$ = $dfrac{f(x)-f(2)}{x-2}$

#### (d)

Here we have that tangent line in $x=2$ is equivalent to the function:

#### (a)

Set A

For the function $f(x)=-x^2+6x-4$ in $x=3$, we can find slope of the tangent in this point as derivative of this function in $x=3$.

So, the derivative of this function is:

$f'(x)=-2x+6$

$textbf{And the slope of tangent in point $x=3$ is}$ $f'(3)=-2cdot3+6=0$

For the function $g(x)=sin{x}$ in $x=3$, we can find slope of the tangent in this point as derivative of this function in $x=dfrac{pi}{2}$.

So, the derivative of this function is:

$g'(x)=cos{x}$

$textbf{And the slope of tangent in point $x=dfrac{pi}{2}$ is}$ $g'(dfrac{pi}{2})=cos{dfrac{pi}{2}}=0$

For the function $h(x)=-x^2+4x+11$ in $x=-2$, we can find slope of the tangent in this point as derivative of this function in $x=-2$.

So, the derivative of this function is:

$h'(x)=2x+4$

$textbf{And the slope of tangent in point $x=-2$ is}$ $h'(-2)=2cdot4+4=0$

For the function $j(x)=5$ in $x=1$, we can find slope of the tangent in this point as derivative of this function in $x=1$.

So, the derivative of this function is:

$j'(x)=0$

$textbf{And the slope of tangent in point $x=1$ is}$ $j'(1)=0$

Set B

For the function $f(x)=3x^2+2x-1$ in $x=2$, we can find slope of the tangent in this point as derivative of this function in $x=2$.

So, the derivative of this function is:

$f'(x)=6x+2$

$textbf{And the slope of tangent in point $x=2$ is}$ $f'(2)=6cdot2+2=14$

For the function $g(x)=2^x+3$ in $x=1$, we can find slope of the tangent in this point as derivative of this function in $x=1$.

So, the derivative of this function is:

$g'(x)=2^xln{2}$

$textbf{And the slope of tangent in point $x=1$ is}$ $g'(1)=2^1ln2=2ln2$

For the function $h(x)=5x+4$ in $x=3$, we can find slope of the tangent in this point as derivative of this function in $x=3$.

So, the derivative of this function is:

$h'(x)=5$

$textbf{And the slope of tangent in point $x=3$ is}$ $h'(3)=5$

For the function $j(x)=sin{x}$ in $x=dfrac{pi}{3}$, we can find slope of the tangent in this point as derivative of this function in $x=dfrac{pi}{3}$.

So, the derivative of this function is:

$j'(x)=cos{x}$

$textbf{And the slope of tangent in point $x=dfrac{pi}{3}$ is}$ $j'(dfrac{pi}{3})=cos{dfrac{pi}{3}}=dfrac{1}{2}$

Set C

For the function $f(x)=3x^2+2x-1$ in $x=-1$, we can find slope of the tangent in this point as derivative of this function in $x=-1$.

So, the derivative of this function is:

$f'(x)=6x+2$

$textbf{And the slope of tangent in point $x=-1$ is}$ $f'(-1)=6cdot({-1})+2=-4$

For the function $g(x)=-2^x+3$ in $x=0$, we can find slope of the tangent in this point as derivative of this function in $x=0$.

So, the derivative of this function is:

$g'(x)=-2^xln2$

$textbf{And the slope of tangent in point $x=0$ is}$ $g'(0)=-2^0ln2=-ln2$

For the function $h(x)=-3x+5$ in $x=2$, we can find slope of the tangent in this point as derivative of this function in $x=2$.

So, the derivative of this function is:

$hf'(x)=-3$

$textbf{And the slope of tangent in point $x=2$ is}$ $h'(2)=-3$

For the function $j(x)=sin{x}$ in $x=2dfrac{pi}{3}$, we can find slope of the tangent in this point as derivative of this function in $x=2dfrac{pi}{3}$.

So, the derivative of this function is:

$j'(x)=cos{x}$

$textbf{And the slope of tangent in point $x=2dfrac{pi}{3}$ is}$ $j'(2dfrac{pi}{3})=cos{2dfrac{pi}{3}}=-dfrac{1}{2}$

#### (b)

Set A

All the slopes of the tangents at this set are $textbf{equivalent to}$ $0$.

Set B

All the slopes of the tangents at this set are $textbf{positive.}$

Set C

All the slopes of the tangents at this set are $textbf{negative.}$

#### (a)

Here we have $textbf{graph}$ of given data:

#### (b)

Here we have $textbf{a curve of best fit and the tangent line at}$ $x=5$:

#### (c)

We have that the slope of the tangent line in $x=5$ is $32.47$.We got this value using quadriatic regression to estimate given data and derivative in $x=5$.

#### (d)

Here we have that $textbf{the instantaneous rate of change in temperature}$ at
exactly 5 min using a centred interval from the table of values is:

$dfrac{Delta{T}}{Delta{x}}=dfrac{310-250}{6-4}=30$

#### (e)

We have that slope of tangent line is a little bit bigger than instaneous rate of change in temperature in same point, $x=5$.

Here we have **the graph of the function** $f(x)=x^2$:

subsection*{(a)}

Here we have graph of this function and the line that passes through $(1,2)$ and $(2,4)$:

subsection*{(b)}

**c)** We know that the first derivative of a function $f(x)$ denotes the slope of the tangent to that function at any point $x$. This means that:

$$
f'(x)=2x,
$$

is the function of the slope of the tangent of the curve $y=x^2$ at any point $x$.

Now, from part b), we know that the slope of the tangent line is $2$. Substituting in the tangent slope equation, we get:

$$begin{aligned}
2x&=2\
x&=1.
end{aligned}$$

Finally, we know that the point of tangency must lie on the curve, which means we can substitute the $x$ coordinate in the equation of the curve to obtain the $y$ coordinate, i.e.,

$$
y=x^2=1^2=1.
$$

Therefore, we can conclude that the point of tangency with the same slope as the secant line is at $(1,1)$.