Determine the maximum possible number of turning points for the graph of the function. f(x) = x^2+7x^3
a polynomial of degree n has at most n-1 turning points
Well, I could try to count them, but I might end up needing an extra pair of hands and maybe a few extra noses to keep track! The thing is, this function has a degree of 3, which means it's a cubic polynomial. A cubic function can have up to two turning points. So let's keep it simple and go with the maximum of two turning points for this one!
To determine the maximum possible number of turning points for the graph of the function f(x) = x^2 + 7x^3, we need to look at the behavior of the second derivative.
First, let's find the first derivative of f(x) by taking the derivative of each term separately:
f'(x) = d/dx(x^2) + d/dx(7x^3)
= 2x + 21x^2
Next, let's find the second derivative by taking the derivative of f'(x):
f''(x) = d/dx(2x + 21x^2)
= 2 + 42x
The number of turning points occurs where the second derivative changes sign. This means that we need to find when f''(x) = 0.
Setting f''(x) = 0, we have:
2 + 42x = 0
42x = -2
x = -2/42
x = -1/21
Since the second derivative is a linear function, it can only change sign at most once. Therefore, the maximum possible number of turning points for the graph of the function f(x) = x^2 + 7x^3 is 1.
To determine the maximum possible number of turning points for the graph of the function f(x) = x^2 + 7x^3, we need to understand what turning points are and how they relate to the given function.
A turning point, also known as a local extremum, is a point on the graph of a function where the function changes from increasing to decreasing or vice versa. It is characterized by the slope of the graph changing from positive to negative or from negative to positive.
In order to find the turning points of a function, we need to find the critical points. Critical points are the values of x for which the derivative of the function is equal to zero or is undefined.
Let's find the derivative of f(x) to find the critical points:
f(x) = x^2 + 7x^3
To find the derivative, we apply the power rule for differentiation. The power rule states that d/dx(x^n) = n*x^(n-1).
Taking the derivative of f(x), we get:
f'(x) = d/dx(x^2) + d/dx(7x^3)
= 2x + 21x^2
To find the critical points, we set the derivative equal to zero and solve for x:
2x + 21x^2 = 0
Factoring out x, we get:
x(2 + 21x) = 0
So, we have two critical points: x = 0 and x = -2/21.
Now, let's analyze the nature of these critical points by evaluating the second derivative of f(x). The second derivative helps us determine whether a critical point corresponds to a maximum, minimum, or an inflection point.
To find the second derivative, we differentiate the first derivative:
f''(x) = d/dx(2x + 21x^2)
= 2 + 42x
Now, let's evaluate the second derivative at the critical points we found earlier:
For x = 0:
f''(0) = 2 + 42(0) = 2
For x = -2/21:
f''(-2/21) = 2 + 42(-2/21) = 2 - 4 = -2
From the second derivative test, we observe that at x = 0, the second derivative is positive (2), indicating a local minimum. At x = -2/21, the second derivative is negative (-2), suggesting a local maximum.
Therefore, the function f(x) = x^2 + 7x^3 has a maximum of 1 turning point, which occurs at x = -2/21.