Near x=0, the tangent line approximation gives e^(−3x)
To find the tangent line approximation near x=0 for the function f(x)=e^(-3x), we can use the concept of derivatives.
The tangent line to a curve at a given point represents the best linear approximation of the curve near that point. The slope of the tangent line is equal to the derivative of the function at that point.
Let's find the derivative of f(x)=e^(-3x). The derivative of e^(-3x) can be found using the chain rule of differentiation. The chain rule states that if we have a composite function, f(g(x)), then the derivative is given by the product of the derivative of the outer function evaluated at the inner function, and the derivative of the inner function.
In this case, g(x)=-3x and f(x)=e^x. The derivative of f(x) with respect to g(x) is e^g(x), and the derivative of g(x) with respect to x is -3. Therefore, applying the chain rule, the derivative of f(g(x)) with respect to x is e^g(x) * (-3).
So, the derivative of f(x)=e^(-3x) is f'(x) = e^(-3x) * (-3).
Now, let's evaluate the derivative at x=0 to find the slope of the tangent line at that point. Substituting x=0 into the derivative equation, we have f'(0) = e^(-3 * 0) * (-3) = -3.
Since the slope of the tangent line is -3, we can write the equation of the tangent line using the point-slope form: y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency, which in this case is (0, e^0) = (0, 1).
Therefore, the equation of the tangent line approximation near x=0 for the function f(x)=e^(-3x) is y - 1 = -3(x - 0), which simplifies to y = -3x + 1.