Final answer:
By taking the derivative of the given function y = 2e^x + 3x + 5x^3, we derived that y' = 2e^x + 3 + 15x^2, which represents the slope of the tangent line at any point x. Setting this equal to 2 and attempting to solve, we end up with 2e^x + 1 + 15x^2 = 0, a complex equation without an algebraic solution. With the quadratic term dominating for larger x values, it is conclusive that there is no tangent line with slope 2.
Step-by-step explanation:
We are asked to show that the curve y = 2ex + 3x + 5x3 has no tangent line with a slope of 2. To do this, let's find the derivative of the curve, which will give us the slope of the tangent line at any given point on the curve. Starting with the original function y = 2ex + 3x + 5x3, taking the derivative with respect to x gives us y' = 2ex + 3 + 15x2. This derivative represents the slope of the tangent line at any point x on the curve. To check if the tangent line can have a slope of 2, we set the derivative equal to 2 and solve for x: 2ex + 3 + 15x2 = 2. Subtracting 2 from both sides gives us 2ex + 3 + 15x2 - 2 = 0, simplifying to 2ex + 1 + 15x2 = 0. This is an exponential and quadratic equation mixed together, which typically does not have an algebraic solution. Based on the nature of the function parts (exponential and quadratic), for larger values of x, the term 15x2 will significantly outweigh the other terms, meaning the derivative (thus the slope) will be much larger than 2. Hence, there is no value of x that will make the derivative equal to 2, confirming that there is no tangent line with a slope of 2.