Find the equations for the tangent line and normal line to the curve for f(x)=(x^4)+(2e^x) at the point (0,2)?
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To find the equations for the tangent and normal lines at a given point on a curve, we need to calculate the derivative of the function at that point.
Let's start by finding the derivative of the function f(x) = x^4 + 2e^x using the sum/difference and chain rules of differentiation.
The derivative of x^4 is 4x^3, and the derivative of 2e^x is 2e^x since the derivative of e^x is e^x itself.
So, the derivative of f(x) = x^4 + 2e^x is f'(x) = 4x^3 + 2e^x.
Now, let's find the slope of the tangent line by evaluating the derivative at the given point (0, 2).
Plug in x = 0 into the derivative function: f'(0) = 4(0)^3 + 2e^0 = 0 + 2(1) = 2.
Therefore, the slope of the tangent line is 2.
Since the tangent line and normal line will have the same slope (since they are perpendicular), the slope of the normal line is also 2.
We have the slope (m = 2) and a point (0, 2). Now we can find the equations for the tangent and normal lines using the point-slope form.
Equation of the tangent line:
Using the point-slope form: y - y₁ = m(x - x₁),
where (x₁, y₁) is the given point (0, 2), and m is the slope of the tangent line (m = 2), we get:
y - 2 = 2(x - 0).
Simplifying, we get:
y - 2 = 2x.
Equation of the normal line:
Using the same point-slope form: y - y₁ = m(x - x₁),
where (x₁, y₁) is the given point (0, 2), and m is the slope of the normal line (m = 2), we get:
y - 2 = 2(x - 0).
Simplifying, we get:
y - 2 = 2x.
Therefore, the equations for both the tangent line and normal line to the curve of f(x) = x^4 + 2e^x at the point (0, 2) are y - 2 = 2x.