find equations of the tangent line and normal line to the curve at the given point. y = x^4 + 3e^x, (0, 3)…

Explanation:

Step1: Find the derivative of the function

The derivative of $y = x^{4}+3e^{x}$ using the sum - rule and power - rule and the derivative of the exponential function. The derivative of $x^{n}$ is $nx^{n - 1}$ and the derivative of $e^{x}$ is $e^{x}$. So $y'=4x^{3}+3e^{x}$.

Step2: Evaluate the derivative at the given point

Substitute $x = 0$ into $y'$. When $x = 0$, $y'(0)=4(0)^{3}+3e^{0}=3$. The slope of the tangent line $m_{t}$ at the point $(0,3)$ is 3.

Step3: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,3)$ and $m = 3$. So $y-3 = 3(x - 0)$, which simplifies to $y=3x + 3$.

Step4: Find the slope of the normal line

The slope of the normal line $m_{n}$ is the negative reciprocal of the slope of the tangent line. Since $m_{t}=3$, then $m_{n}=-\frac{1}{3}$.

Step5: Find the equation of the normal line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(0,3)$ and $m=-\frac{1}{3}$. So $y - 3=-\frac{1}{3}(x - 0)$, which simplifies to $y=-\frac{1}{3}x+3$.

Answer:

tangent line: $y = 3x + 3$ normal line: $y=-\frac{1}{3}x + 3$