Calculus - stationary points and functions

Question

Consider the function f(x) = x^4 + 2x^2 - 3.

(a) Use calculus to find the stationary points of f. For each stationary point,

determine whether it is a maximum, minimum or point of in

(b) Use the substitution u = x2 and give an expression for f(u). Use the resulting

quadratic equation to find where f intercepts the x-axis.

c) State the coordinates of the intersection of f and the y-axis.

(d) Is f a continuous function? Explain why.

(e) Sketch the function and verify your answers to (a) and (c).

Answer 1

(a) For this question you want to find where the derivative is zero or undefined. When you take the derivative of f(x) you get f'(x) = 4x³ + 4x. There aren't any x values where this function is undefined, so you can set f'(x) equal to 0. After factoring out 4x you then get 4x(x² + 1) = 0. You can then set each individual factor equal to 0. x² + 1 = 0 won't have a rational number answer, but 4x = 0 gives you x = 0 as a stationary point.

To figure out whether this is a maximum or a minimum there are two ways to do this. If you've learned second derivatives, then you can take the second derivative of f(x) and plug in x = 0. If the second derivative at x = 0 is positive, then the stationary point is a minimum. If the second derivative is negative, then it's a maximum.

The other way is to look at the values for f'(x) on either side of the stationary point. You can do this by picking any number to the left and right of the stationary point. For simplicity's sake I'd recommend 1 and -1 for this problem. If the value of f'(x) that's to the left of x = 0 is negative and the value to the right of x = 0 is positive, then x = 0 is a minimum. If the value to the left is positive and the value to the right is negative, then x = 0 is a maximum.

(b) For this you want to replace every x² with u. This gives you f(u) = u² + 2u - 3. To find where f(u) intercepts the x-axis you want to set f(u) equal to zero. After doing this and factoring the quadratic equation you get (u + 3)(u - 1) = 0. Solve this to get u = -3 and u = 1. This means that f(u) intercepts the x-axis at u = -3 and u = 1. To find the x-values you need to plug the u-values into u = x². It's easier to change the equation first to x = √u. Plugging the u values in you get x = ±√-3 and x = ±√1. Since the first solution doesn't produce a real number you can ignore it and just use the second solution, where f intercepts the x-axis at x = 1 and x = -1.

(c) Here you just plug in x = 0 into f(x), and you get f(0) = -3, so f intercepts the y-axis at y = -3.

(d) Yes because all polynomials are continuous functions.

Answer 2

(a) To find the stationary points of f(x) = x^4 + 2x^2 - 3, we need to find the values of x where the derivative of f(x) is equal to 0 or does not exist.

The derivative of f(x) = x^4 + 2x^2 - 3 is f'(x) = 4x^3 + 4x

Setting f'(x) = 0, we get 4x^3 + 4x = 0, which simplifies to x(x^2 + 1) = 0. Therefore, the stationary points of f(x) are x = 0 and x = ±i.

To determine whether this stationary point is a maximum, minimum, or point of inflection, we need to look at the second derivative of f(x).

The second derivative of f(x) is f''(x) = 12x^2 + 4

We can see that f''(x) is positive for all x, which means that the stationary point x = 0 is a minimum.

(b) By substituting u = x^2 into f(x) = x^4 + 2x^2 - 3, we get f(u) = u^2 + 2u - 3.

This is a quadratic equation in the form of f(u) = au^2 + bu + c. The x-intercepts are the zeros of the function, which occur when f(u) = u^2 + 2u - 3 = (u+3)(u-1) = 0

Then u + 3 = 0 and u - 1 = 0.

Thus, x^2 + 3 = 0, giving x +/-sqrt(-3), and x^2 - 1 = 0, giving x = +/-1. Our x-intercepts are thus (-1,0) and (1,0).

(c) The y-intercept of a function is the point at which the function crosses the y-axis, which occurs when x = 0. Therefore, the coordinates of the intersection of f and the y-axis are (0, -3)

(d) A function is continuous if the limit of the function as x approaches a particular value exists and is equal to the function evaluated at that value.

The function f(x) = x^4 + 2x^2 - 3 is a polynomial function, which is a continuous function on the entire real line.

(e) I don't seem to be able to insert images, but the graph of f(x) is a parabola that opens upwards with a its minimum point/vertex at (0,-3).