Find the Maximum/Minimum Value 6x^4-32x^3+48x^2-7
This problem is asking for the determination of the extremum (either the maximum or minimum value) of a given fourth-degree polynomial function f(x) = 6x^4 - 32x^3 + 48x^2 - 7. To solve this, one would typically calculate the first derivative of the function to find the critical points, then apply the second derivative test or analyze the first derivative sign changes around the critical points to determine the nature of each extremum (whether it is a local maximum, local minimum, or neither). The global maximum or minimum would then be found by comparing these local extrema with the function values at the endpoints or as the variable x approaches infinity, if the domain is unbounded.
$6 x^{4} - 32 x^{3} + 48 x^{2} - 7$
Answer
Solution:
Step 1: Derive the function to find critical points
Take the derivative of $f(x) = 6x^4 - 32x^3 + 48x^2 - 7$.
Apply the Power Rule: $\frac{d}{dx}x^n = nx^{n-1}$.
Derivative: $f'(x) = 24x^3 - 96x^2 + 96x$.
Step 2: Find the second derivative for the concavity test
Derive $f'(x)$ to get $f''(x)$.
Second Derivative: $f''(x) = 72x^2 - 192x + 96$.
Step 3: Solve $f'(x) = 0$ to find critical points
Factor $f'(x)$: $24x(x^2 - 4x + 4) = 24x(x - 2)^2$.
Set equal to zero: $x = 0, 2$.
Step 4: Determine the nature of critical points
Use the second derivative test.
For $x = 0$: $f''(0) = 96 > 0$, so it's a local minimum.
For $x = 2$: $f''(2) = 0$, inconclusive, use the first derivative test.
Step 5: Apply the first derivative test for $x = 2$
Check sign changes of $f'(x)$ around $x = 2$.
No sign change implies $x = 2$ is neither a maximum nor a minimum.
Step 6: Find the y-values for the critical points
For $x = 0$: $f(0) = -7$.
The local minimum is at $(0, -7)$.
Step 7: Conclusion
The function has a local minimum at $(0, -7)$.
No local maximum found.
Knowledge Notes:
Power Rule: Used to differentiate terms with powers of $x$. If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.
Product Rule: Used when differentiating products of functions. If $u(x)$ and $v(x)$ are functions of $x$, then the derivative of $u(x)v(x)$ is $u'(x)v(x) + u(x)v'(x)$.
Sum Rule: Used when differentiating a sum of functions. The derivative of a sum is the sum of the derivatives.
Second Derivative Test: If $f''(x) > 0$, the function is concave up and the point is a local minimum. If $f''(x) < 0$, it's concave down and the point is a local maximum.
First Derivative Test: Used to determine if a critical point is a local maximum or minimum by analyzing the sign changes of the first derivative around the critical point.
Critical Points: Points where the first derivative is zero or undefined. They are potential locations for local maxima, minima, or points of inflection.
Solution:"The function $f(x) = 6x^4 - 32x^3 + 48x^2 - 7$ has a local minimum at the point $(0, -7)$. No local maximum was found." Knowledge Notes:"The solution involves applying calculus concepts such as the Power Rule, Product Rule, Sum Rule, and the First and Second Derivative Tests to determine the local extrema of the function."
