How do I find the maxima and minima of f(x) = e^(x^2)?

When dealing with maxima and minima points, there are two ways to go, one of these is to first compute the first derivative of the function, check when this function is zero, and then study its sign; the other is to check the second derivative. We'll go with the first method here.We first compute the first derivative of our function f; we notice that we have a function of the form g(h(x)), where g(x) = e^x, and h(x) = x^2, thus we may use the chain rule to differentiate:f'(x) = e^(x^2) * 2x We now check when the function f(x) = 0; we notice that this function is zero when either e^(x^2) or 2x is zero. We know that the exponential function is never zero, thus this function is zero if and only if 2x is zero, i.e. when x=0. To check if this corresponds to a point of maximum or minimum, we need to check the sign of the first derivative by checking the inequality e^(x^2) * 2x > 0.The exponential function is always positive, thus the inequality can be solved by solving the inequality 2x>0. By this and the information we got above we get that:f'(x) > 0 when x > 0f'(x) < 0 when x < 0f'(x) = 0 when x = 0The first derivative tells us how the function changes over the domain (the gradient) by telling us the slope of the tangent line at each point. The gradient of our graph is negative (the tangent points down) before 0 and positive (the tangent points up) after 0, thus our point x=0 corresponds to a point of minimum of our function (the graph "decreases" before it and "increases" after it).We can now compute f(0) = 1, giving the minimum (0, 1).