Where does f(x)=x3+3x2−5x−2 have a local maximum? (1 point) Responses x=1+26√3 x = 1 + 2 6 3 x=−1−6√3 x = − 1 − 6 3 x=−1+6√3 x = − 1 + 6 3 x=−1−26√3 x = − 1 − 2 6 3 x=−1+26√3 x = − 1 + 2 6 3 Question 2 Part B What is the x-value of the inflection point of f(x) ? (1 point) Responses x=−1 x = − 1 x=−12 x = − 1 2 x = 0 x = 0 x=12 x = 1 2 x = 1 x = 1 Skip to navigation
To find the local maximum of the function f(x) = x^3 + 3x^2 - 5x - 2, we need to find the critical points by taking the derivative of the function and setting it to zero.
f'(x) = 3x^2 + 6x - 5
Setting f'(x) = 0 gives us:
3x^2 + 6x - 5 = 0
Using the quadratic formula, we find the solutions for x:
x = (-6 ± √(6^2 - 4*3*(-5))) / (2*3)
x = (-6 ± √(36 + 60)) / 6
x = (-6 ± √96) / 6
x = (-6 ± 4√6) / 6
x = -1 ± (2/3)√6
So, the critical points are x = -1 + (2/3)√6 and x = -1 - (2/3)√6.
To find the local maximum, we need to test the values between and around the critical points. Taking the second derivative:
f''(x) = 6x + 6
and substituting the values of the critical points into f''(x), we get:
f''(-1 + (2/3)√6) = 6(-1 + (2/3)√6) + 6 = -4 + 4√6
f''(-1 - (2/3)√6) = 6(-1 - (2/3)√6) + 6 = -4 - 4√6
Since the coefficient of x^2 in f''(x) is positive, we know that if the second derivative is positive at a point, then it is a local minimum, and if the second derivative is negative, then it is a local maximum.
From the second derivative tests, we can conclude that the local maximum of f(x) occurs at x = -1 + (2/3)√6.
For the inflection point of f(x), we need to find where the concavity changes. This occurs when the second derivative changes sign.
Since f''(x) = 6x + 6 is a linear function, it does not change signs. Therefore, there is no inflection point in the graph of f(x).