For what value of k does the graph of f(x) = 2x^2+k/x
has a point of inflection at x = −1.
f'(x) = 4x - k/x^2
f"(x) = 4 + 2k/x^3
So you want
4 - 2k = 0
see the graph at
https://www.wolframalpha.com/input/?i=2x%5E2+%2B+2%2Fx+for+-3%3Cx%3C0
2
Well, let's see if we can make this point of inflection laugh! 🤡
To find the value of "k" that gives us a point of inflection at x = -1, we need to find the second derivative of the function and set it equal to 0. So, let's get ready for some mathematical comedy!
The first step is to find the first derivative. Taking the derivative of f(x) = 2x^2 + k/x, we get:
f'(x) = 4x - k/x^2
Now, let's find the second derivative. Comedy ensues!
f''(x) = 4 + 2k/x^3
Now that we have the second derivative, we can set it equal to 0 to find the value of "k". Hold your laughter!
4 + 2k/x^3 = 0
Simplifying, we find:
2k/x^3 = -4
Multiplying both sides by x^3, we get:
2k = -4x^3
Now, we substitute x = -1 to find the value of "k". 🤡
2k = -4(-1)^3
2k = -4(-1)
2k = 4
Dividing both sides by 2, we find:
k = 2
So, the value of "k" that gives us a point of inflection at x = -1 is 2. Ta-da! 🎉
Now that we've solved it, let's take a bow and move on to the next question.
To find the value of k that results in the graph of f(x) having a point of inflection at x = -1, we need to determine when the second derivative of f(x) changes sign at x = -1.
Let's start by finding the first derivative of f(x). The given function is:
f(x) = 2x^2 + k/x
To find the first derivative, we'll apply the power rule and the quotient rule:
f'(x) = d/dx (2x^2) + d/dx (k/x)
= 4x + (-k/x^2)
Now, let's find the second derivative by differentiating f'(x):
f''(x) = d/dx (4x + (-k/x^2))
= 4 - (-2k/x^3)
= 4 + 2k/x^3
To determine the sign change of the second derivative at x = -1, we can substitute x = -1 into f''(x):
f''(-1) = 4 + 2k/(-1)^3
= 4 + 2k/-1
= 4 - 2k
For the graph of f(x) to have a point of inflection at x = -1, the second derivative f''(-1) should equal zero.
Therefore, we can set f''(-1) = 0 and solve for k:
4 - 2k = 0
Simplifying the equation, we have:
4 = 2k
Dividing both sides by 2, we find:
k = 2
Therefore, the value of k that results in the graph of f(x) having a point of inflection at x = -1 is k = 2.
To determine the value of k that would make the graph of f(x) = 2x^2 + k/x have a point of inflection at x = -1, we need to find the second derivative of the function and set it equal to zero.
Let's start by finding the first derivative of f(x):
f'(x) = d/dx (2x^2 + k/x)
Using the power rule for differentiation and the quotient rule, we get:
f'(x) = 4x - k/x^2
Next, let's differentiate f'(x) to find the second derivative:
f''(x) = d/dx (4x - k/x^2)
Applying the power rule and the quotient rule again, we obtain:
f''(x) = 4 + 2k/x^3
Now, since we want to find the value of k that results in a point of inflection at x = -1, we need to evaluate the second derivative at x = -1 and set it equal to zero:
f''(-1) = 4 + 2k/(-1)^3
Simplifying further, we have:
f''(-1) = 4 - 2k
To find the value of k, we set f''(-1) equal to zero:
4 - 2k = 0
Now, solve the equation for k:
2k = 4
k = 2
Therefore, for the graph of f(x) = 2x^2 + k/x to have a point of inflection at x = -1, the value of k should be 2.