Question

find the slope of the graph of the function at the given point. use proper notation.
5.) (f(\theta)=4sin\theta - \theta), ((0,0))
(f(\theta)=4cos\theta - 1)
(f(0)=4cos(0)-1 = 3)
6.) (f(x)=\frac{3}{4}e^{x}), ((0,\frac{3}{4}))
(f(x)=\frac{3}{4}e^{x})
(f(0)=\frac{3}{4}e^{(0)}=\frac{3}{4})
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))
(f(x)=1 + e^{x})
(f(0)=1+e^{(0)} = 2)
(y - 1=2(x - 0))
8.) (g(t)=sin t+\frac{1}{2}e^{t}), ((pi,\frac{1}{2}e^{pi}))
(g(t)=cos t+\frac{1}{2}e^{t})
(g(0)=cos(pi)+\frac{1}{2}e^{(pi)}=)

Explanation:

Step1: Recall derivative - slope relationship

The slope of the graph of a function $y = f(x)$ at a point $(x_0,y_0)$ is given by $f^{\prime}(x_0)$.

Step2: Differentiate $g(t)$

Given $g(t)=\sin t+\frac{1}{2}e^{t}$, by the sum - rule of differentiation and the derivatives of basic functions ($(\sin t)^{\prime}=\cos t$ and $(e^{t})^{\prime}=e^{t}$), we have $g^{\prime}(t)=\cos t+\frac{1}{2}e^{t}$.

Step3: Evaluate $g^{\prime}(t)$ at $t = \pi$

Substitute $t=\pi$ into $g^{\prime}(t)$. We know that $\cos(\pi)=- 1$ and $e^{\pi}$ is a non - zero constant. So $g^{\prime}(\pi)=\cos(\pi)+\frac{1}{2}e^{\pi}=-1+\frac{1}{2}e^{\pi}$.

Step4: Write the equation of the tangent line

The point - slope form of a line is $y - y_0=m(x - x_0)$. Here $x_0=\pi$, $y_0=\frac{1}{2}e^{\pi}$, and $m = g^{\prime}(\pi)=-1+\frac{1}{2}e^{\pi}$. So the equation of the tangent line is $y-\frac{1}{2}e^{\pi}=(-1 + \frac{1}{2}e^{\pi})(t-\pi)$.

Answer:

The slope of the graph of $g(t)$ at $t = \pi$ is $-1+\frac{1}{2}e^{\pi}$, and the equation of the tangent line is $y-\frac{1}{2}e^{\pi}=(-1+\frac{1}{2}e^{\pi})(t - \pi)$