How can the graph y = x sqaure root(9-2x squared)with the point (3root2/root 2,0) have in infinite gradient?
To determine whether the graph of a function has an infinite gradient at a particular point, we need to examine the behavior of the derivative at that point. If the derivative approaches infinity as we approach the point from both sides, then the graph will have an infinite gradient at that point.
Let's start by finding the derivative of the function. The given function is y = x * √(9 - 2x^2). To find the derivative, we can use the product rule and the chain rule.
First, let's rewrite the function as y = x(9 - 2x^2)^(1/2).
Using the product rule, the derivative of y with respect to x can be determined as:
dy/dx = (9 - 2x^2)^(1/2) + x * (1/2)(9 - 2x^2)^(-1/2)(-4x).
Next, we need to assess the behavior of this derivative as x approaches the given point (3√2 / √2, 0).
Substituting the x-coordinate of the point into the derivative expression, we have:
dy/dx = (9 - 2*(3√2 / √2)^2)^(1/2) + (3√2 / √2) * (1/2)(9 - 2*(3√2 / √2)^2)^(-1/2)(-4(3√2 / √2)).
Simplifying this expression:
dy/dx = (9 - 18/2)^(1/2) + (3√2 / √2) * (1/2)(9 - 18/2)^(-1/2)(-4(3√2 / √2)).
dy/dx = (9 - 9)^(1/2) + (3√2 / √2) * (1/2)(9 - 9)^(-1/2)(-4(3√2 / √2)).
dy/dx = 0 * (1/2)(0)^(-1/2)(-4(3√2 / √2)).
From the derivative expression, it is clear that the derivative is 0 at the point (3√2 / √2, 0). Therefore, the graph of the function does not have an infinite gradient at that point.
It is important to note that the graph may have vertical tangent lines at other points, but not at the point in question.