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}, (\frac{\pi}{2},\frac{1}{2 - e^{\pi}})

Explanation:

Step1: Find the derivative of the function

For \(y = f(x)=x + e^{x}\), using the sum - rule and the derivative formulas \(\frac{d}{dx}(x)=1\) and \(\frac{d}{dx}(e^{x})=e^{x}\), we have \(y'=f'(x)=1 + e^{x}\).

Step2: Evaluate the derivative at the given \(x\) - value

The point is \((0,1)\), when \(x = 0\), \(f'(0)=1+e^{0}=1 + 1=2\). The slope \(m\) of the tangent line is \(2\).

Step3: Use the point - slope form of a line

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

Step4: Simplify the equation

\(y-1=2x\), so \(y = 2x+1\).

Answer:

\(y = 2x + 1\)