Find the x-value where the relative maximum of y = e^x/5−x^2 closest to x = 0 occurs.

Question

When you need to solve for an expression equal to zero, use Newton's method with  as your first guess and enter your second guess as the answer. Enter your answer as a fraction.

Answer 1

f(x) = e^x/5 - x²; the domain of this function is all real numbers. To determine relative max and min set the 1st derivative equal to zero.

f'(x) = e^x/5 - 2x

e^x/5 - 2x = 0; this equation cannot be solved algebraically. The post suggests to use Newton's Method with a starting value x₀ = 0.

Let's define f'(x) as g(x) = e^x/5 - 2x

g'(x) = e^x/5 - 2

N(x) = x - [g(x)/g'(x)] is the function for Newton's Method.

x₁ = N(x₀) = N(0) = 0 - [g(0)/g'(0)] = - [(1/5)/((1/5) - 2)] = 1/9.

Of course you could continue:

x₂ = N(x₁) = N(1/9) ≈ 0.1118325

Turns out this is the only x-value where a relative max occurs.

g(x) also equals zero at approximately: 3.577152. A relative min occurs at that x-value.

This Desmos graph shows the original function along with a graph of f'(x). It also shows a few more iterations of Newton's Method:

desmos.com/calculator/y0az95jmyf