First, take the derivative of f(x). This is very easy:

f'(x) = 12x² – 4x

Now, the slope of the tangent line at x = 5 is f'(5). Evaluated, this is: f'(5) = 12 * 5 * 5 – 4 * 5 = 300 – 20 = 280

Now, we must find the intersection point on the original line:

f(5) = 4 * 5³ – 2 * 5² + 4 = 500 – 50 + 4 = 454

Therefore, the point of tangent intersection is (5,454).

Using the point-slope form of linear equations, we can find the line:

(y – 454) = 280(x – 5)

y – 454 = 280x – 1400

y = 280x – 946

The equation of the tangent line will have the form $\displaystyle y-y_{0}=m(x-x_{0})$, where $\displaystyle m$ is the slope of the line and $\displaystyle (x_{0},y_{0})=(1,3)$.  

To find the slope, we need to evaluate the derivative at $\displaystyle x=1$:

 $\displaystyle \frac{dy}{dx}=\frac{d(x^2-x+3)}{dx}=2x-1\Rightarrow m=f'(1)=2*1-1=1$

Now that we have the slope, we can determine the equation of the tangent line using the point-slope formula:

 $\displaystyle y-y_{0}=m(x-x_{0})\Rightarrow y-3=1(x-1)\Rightarrow y=x+2.$

To find the equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form $\displaystyle y=mx+b$, where m is the slope and $\displaystyle b$ is the y adjustment. To get the slope, find the derivative of $\displaystyle f(x)$ and plug in the desired point $\displaystyle 0$ for $\displaystyle x$, giving us an answer of $\displaystyle 3$ for the slope.

$\displaystyle f'(x)=3e^{3x} + 2x$

$\displaystyle f'(0)=3$

To find the y adjustment pick a point 0 in the original$\displaystyle f(x)$ function. For simplicity, let's plug in $\displaystyle x=0$, which gives us a y of 1, so an easy point is $\displaystyle (0,1)$. Next plug in those values into the equation of a line, $\displaystyle y=mx + b$. The new equation with all parameters plugged in is

$\displaystyle 1=(3)(0)+b$

Now you simply solve for $\displaystyle b$, which is $\displaystyle 1$.

Final equation of the line tangent to $\displaystyle f(x)$ at $\displaystyle x=0$ is $\displaystyle y=3x+1$

To get the slope, find the derivative of $\displaystyle f(x)$ and plug in the desired point $\displaystyle 2$ for $\displaystyle x$, giving us an answer of $\displaystyle \frac{2}{5}$ for the slope.

$\displaystyle f'(x)=\frac{2}{2x+1}$

$\displaystyle f'(2)=\frac{2}{5}$

Remember that the derivative of $\displaystyle ln(u) = \frac{u'}{u}$.

To find the $\displaystyle y$ adjustment pick a point $\displaystyle 0$ (for example) in the original $\displaystyle f(x)$ function. For simplicity, let's plug in $\displaystyle x=0$, which gives us a $\displaystyle y$ of $\displaystyle 3$, so an easy point is $\displaystyle (0,3)$. Next plug in those values into the equation of a line, $\displaystyle y=mx + b$. The new equation with all parameters plugged in is

$\displaystyle 3=\frac{2}{5}(0)+b$

The coefficient in front of the $\displaystyle 0$ is the slope.

Now you simply solve for $\displaystyle b$, which is $\displaystyle 3$.

Final equation of the line tangent to $\displaystyle f(x)$ at$\displaystyle x=2$ is $\displaystyle y=\frac{2}{5}x +3$.

The equation of the tangent line is $\displaystyle y=mx+b.$ To find $\displaystyle m$, the slope, calculate the derivative and plug in the desired point.

$\displaystyle y'=2cos(2x)$

$\displaystyle y'=2cos\left(2\cdot \frac{\pi}{2}\right)$

$\displaystyle y'=2cos(\pi)$

$\displaystyle y'\left(\frac{\pi}{2} \right )=-2$

The next step is to choose a coordinate on the original $\displaystyle y$ function. We can choose any $\displaystyle x$ value and calculate its $\displaystyle y$ value.

Let's choose $\displaystyle \frac{\pi }{2}$.

$\displaystyle y=sin\left(2\cdot \frac{\pi}{2}\right)=sin(\pi)=0$

The $\displaystyle y$ value at this point is $\displaystyle 0$.

Plugging in those values we can solve for $\displaystyle b$.

$\displaystyle 0=(-2)\left(\frac{\pi }{2}\right)+b$

Solving for $\displaystyle b$ we get $\displaystyle b$=$\displaystyle \pi$.

$\displaystyle y=-2x+\pi$

First we need to find the slope of $\displaystyle f$ at $\displaystyle x=\frac{\pi}{2}$. To do this we need the derivative of $\displaystyle f$. To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.

$\displaystyle \frac{df}{dx}=2x-cos(x)$

At,

 $\displaystyle x=\frac{\pi}{2}, \frac{df}{dx}=\pi$

Now we need to know

$\displaystyle f\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}-1$.

Now we have a slope, $\displaystyle m=\pi$ and a point $\displaystyle (x_1,y_1)=\left(\frac{\pi}{2},\frac{\pi^2}{4}-1\right)$

so we can use the point-slope formula to find the equation of the line.

$\displaystyle y-y_1=m(x-x_1)$

Plugging in and rearranging we find

$\displaystyle y=\pi x-\frac{\pi^2}{4}-1$.

First, evaluate $\displaystyle f(x)$ when $\displaystyle x=1$.

$\displaystyle f(1)=2(1)^4+e^1-e(1)=2+0\textbf{=2}$

Thus, we need a line that contains the point $\displaystyle (1,2)$

Next, find the derivative of $\displaystyle f(x)$ and evaluate it at $\displaystyle x=1$.

To find the derivative we will use the power rule,

$\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f'(x)=8x^3+e^x-e$

$\displaystyle f'(1)=8(1)^3+e^1-e=8+0=8$

This indicates that we need a line with a slope of 8.

In point-slope form, $\displaystyle y-y_{1}=m(x-x_{1})$, a line with the point $\displaystyle (1,2)$ and a slope of 8 will be:

$\displaystyle y-2=8(x-1)$

The tangent line to $\displaystyle f(x)$ at $\displaystyle x=-1$ must have the same slope as $\displaystyle f(x)$.

Applying the chain rule we get

$\displaystyle f'(x)=\frac{1}{1+(1-x^3)^2}(-3x^2)$.

Therefore the slope of the line is, 

$\displaystyle f'(-1)=-\frac{3}{5}=-0.6$.

In addition, the tangent line touches the graph of $\displaystyle f(x)$ at $\displaystyle x=-1$. Since $\displaystyle f(-1)=1.11$, the point $\displaystyle (-1, 1.11)$ lies on the line.

Plugging in the slope and point we get $\displaystyle y=-0.6x+0.51$.

In order to find the equation of the tangent line at $\displaystyle x=1$, we first find the slope.

To do this we need to find $\displaystyle f'(x)$.

$\displaystyle f(x)=x^4-x^2+x-5$

$\displaystyle f'(x)=4x^3-2x+1$

Since we have found $\displaystyle f'(x)$, now we simply plug in 1.

$\displaystyle f'(1)=4(1)^3-2(1)+1=3$

Now we need to plug in 1, into $\displaystyle f(x)$, to find a point that the tangent line touches.

$\displaystyle f(1)=(1)^4-(1)^2+1-5=-4$

Now we can use point-slope form to figure out what the equation of the tangent line is at $\displaystyle x=1$.

Remember that point-slope for is

$\displaystyle y-y_0=m(x-x_0)$

where $\displaystyle x_0$ and $\displaystyle y_0$ is the point where the tangent line touches $\displaystyle f(x)$, and $\displaystyle m$ is the slope of the tangent line.

In our case, $\displaystyle x_0=1$, $\displaystyle y_0=-4$, and $\displaystyle m=3$.

$\displaystyle y+4=3(x-1)$

$\displaystyle y+4=3x-3$

Thus our tangent line equation at $\displaystyle x=1$ is

$\displaystyle y=3x-7$.

In order to find the equation of the tangent line at $\displaystyle x=1$, we first find the slope.

To do this we need to find $\displaystyle f'(x)$ using the power rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(x)=x^3+x$

$\displaystyle f'(x)=3x^2+1$

Since we have found $\displaystyle f'(x)$, now we simply plug in 1.

$\displaystyle f'(1)=3(1)^2+1=4$

Now we need to plug in 1, into $\displaystyle f(x)$, to find a point that the tangent line touches.

$\displaystyle f(1)=(1)^3+1=2$

Now we can use point-slope form to figure out what the equation of the tangent line is at $\displaystyle x=1$.

Remember that point-slope for is

$\displaystyle y-y_0=m(x-x_0)$

where $\displaystyle x_0$ and $\displaystyle y_0$ is the point where the tangent line touches $\displaystyle f(x)$, and $\displaystyle m$ is the slope of the tangent line.

In our case, $\displaystyle x_0=1$, $\displaystyle y_0=2$, and $\displaystyle m=4$.

$\displaystyle y-2=4(x-1)$

$\displaystyle y-2=4x-4$

Thus our tangent line equation at $\displaystyle x=1$ is

$\displaystyle y=4x-2$.