How many critical points does the function:
(x+2)^5 * (x-3)^4
have?
How would I solve this quickly? The answer says something about an effect of multiplicity of the zeros of the function?
A critical point of a function within its domain is any point which is not differentiable or when its derivative is zero.
Since
f(x)=(x+2)^5 * (x-3)^4
is a polynomial, its domain is (-∞,∞) and differentiable throughout.
The only critical points are when the derivative is zero.
f'(x)=0 =>
4(x-3)^3(x+2)^5+5(x-3)^4(x+2)^4=0
which factors to:
(x-3)^3*(x+2)^4*(9*x-7)=0
We see that
x=3 (multiplicity 3)
x=-2 (multiplicity 4)
and
x=7/9
So there is a total of 3+4+1 = 8 critical points, out of which there are 3 distinct points.
Well, let me put on my math clown nose for this one! The number of critical points of a function is determined by the number of changes in concavity. To find the critical points, we need to take the derivative of the function.
So, let's differentiate this function, shall we?
f(x) = (x+2)^5 * (x-3)^4
Now, using the product rule, we get:
f'(x) = 5(x+2)^4 * (x-3)^4 + 4(x+2)^5 * (x-3)^3
To determine the critical points, we need to set the derivative equal to zero and solve for x. However, due to the multiplicity of the zeros, we need to consider the effect on the number of critical points.
The function (x+2)^5 has a zero of multiplicity 5 at x = -2, and the function (x-3)^4 has a zero of multiplicity 4 at x = 3.
So, the effect of multiplicity is that the function has 5 critical points at x = -2 and 4 critical points at x = 3.
I hope my clown math skills were enough to solve this quickly for you!
To find the number of critical points of the function f(x) = (x+2)^5 * (x-3)^4, we first need to understand the concept of critical points.
The critical points of a function occur when its derivative is zero or undefined. In this case, we need to find the derivative of f(x) to determine the critical points.
To calculate the derivative, we can use the product rule. Let's start by finding the derivative of each term separately.
The derivative of (x+2)^5 can be found using the power rule, which states that the derivative of x^n is equal to n * x^(n-1). Applying this rule, we get:
d/dx [(x+2)^5] = 5 * (x+2)^4
Similarly, the derivative of (x-3)^4 can be found as:
d/dx [(x-3)^4] = 4 * (x-3)^3
Now, we can use the product rule to find the derivative of f(x):
f'(x) = [(x+2)^5]' * (x-3)^4 + (x+2)^5 * [(x-3)^4]'
= 5 * (x+2)^4 * (x-3)^4 + 4 * (x+2)^5 * (x-3)^3
To find the critical points, we need to solve the equation f'(x) = 0.
Setting f'(x) = 0, we have:
5 * (x+2)^4 * (x-3)^4 + 4 * (x+2)^5 * (x-3)^3 = 0
Now, to determine the number of critical points, we need to analyze the effect of the multiplicity of the zeros of the function.
From the equation, we can see that either (x+2)^4=0 or (x-3)^4=0. This means that there is a possibility of four critical points at x = -2 and another four at x = 3.
However, we need to consider the multiplicity of the zeros. The multiplicity refers to the number of times a zero occurs.
For example, if the exponent is an even number like 2 or 4, the zero has an even multiplicity. If the exponent is an odd number like 1 or 3, the zero has an odd multiplicity.
In our case, we have (x+2)^4 = 0 and (x-3)^4 = 0, meaning that the zeros at x = -2 and x = 3 each have a multiplicity of 4.
Since the exponents are even, the function touches or crosses the x-axis at these points but does not change sign. Therefore, we classify these points as points of inflection rather than critical points.
Thus, the function f(x) = (x+2)^5 * (x-3)^4 has no critical points.
Therefore, the function f(x) has zero critical points.
To find the number of critical points for the function (x+2)^5 * (x-3)^4, we need to determine the locations where the derivative of the function is zero or undefined.
First, we find the derivative of the function by applying the product rule. The derivative of (x+2)^5 is 5(x+2)^4, and the derivative of (x-3)^4 is 4(x-3)^3.
Next, we set the derivatives equal to zero to find potential critical points:
5(x+2)^4 = 0
4(x-3)^3 = 0
To simplify, we can solve each equation individually:
For 5(x+2)^4 = 0, we have:
(x+2)^4 = 0
Taking the fourth root of both sides, we get:
(x+2) = 0
x = -2
For 4(x-3)^3 = 0, we have:
(x-3)^3 = 0
Taking the cube root of both sides, we get:
(x-3) = 0
x = 3
So, we have found two critical points: x = -2 and x = 3.
Now, let's discuss the effect of multiplicity of the zeros on the critical points. In this function, we have (x+2)^5 and (x-3)^4. The exponent of each term represents the multiplicity of the zero.
A zero with even multiplicity, such as (x-3)^4, means the curve touches the x-axis but does not cross it. So, it does not change the sign of the derivative and does not affect the number of critical points. In this case, the zero x = 3 has multiplicity 4, which is even.
On the other hand, a zero with odd multiplicity, such as (x+2)^5, means the curve crosses the x-axis at that point. It changes the sign of the derivative and contributes to the number of critical points. In this case, the zero x = -2 has multiplicity 5, which is odd.
Therefore, the function (x+2)^5 * (x-3)^4 has a total of two critical points: x = -2 and x = 3.