Final answer:
To find the equation of the line tangent to the graph of f(x) = x^4 + 2x^2 at the point where f'(x) = 1, we find the derivative of f(x), solve for x, find the corresponding y-value, and use the point-slope form to determine the equation of the tangent line.
Step-by-step explanation:
To find the equation of the line tangent to the graph of f(x) = x4 + 2x2 at the point where f'(x) = 1, we need to find the derivative of f(x). Taking the derivative of f(x), we get f'(x) = 4x3 + 4x. Since we are looking for the point where f'(x) = 1, we can set the equation equal to 1 and solve for x: 4x3 + 4x = 1.
Using a calculator or numerical methods, we find that x ≈ 0.226. Now, we can substitute this value of x into the equation of f(x) to find the corresponding y-value: y = (0.226)4 + 2(0.226)2. Evaluating this expression, we get y ≈ 0.543.
Therefore, the equation of the tangent line at the point (0.226, 0.543) is y = mx + b, where m is the slope of the tangent line and b is the y-intercept. Since the slope of the tangent line is equal to f'(x) at the point of tangency, the slope is 1. Using the point-slope form, we have y - 0.543 = 1(x - 0.226), which simplifies to y = x + 0.317.