Question

write the equation of the tangent line to the graph of the function at the given point
7.) $f(x)=x + e^{x}, (0,1)$
8.) $g(t)=sin t+\frac{1}{2}e^{t}, (pi,\frac{1}{2}e^{pi})$

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x + e^{x}$ using the sum - rule and the derivative formulas $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(e^{x})=e^{x}$ is $f'(x)=1 + e^{x}$.

Step2: Evaluate the derivative at the given x - value

We want to find the slope of the tangent line at $x = 0$. Substitute $x = 0$ into $f'(x)$: $f'(0)=1+e^{0}=1 + 1=2$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,1)$ is the given point and $m = 2$ is the slope. Substituting these values, we get $y - 1=2(x - 0)$.

Step4: Simplify the equation

$y-1 = 2x$, so the equation of the tangent line is $y=2x + 1$.

Answer:

$y = 2x+1$