Determine the critical points of the function f(x)=x^4-6x^2+8x+10 and state their nature
To find the critical points of a function, we first need to find its derivative, set it equal to zero, and solve for x.
Taking the derivative of f(x) = x^4 - 6x^2 + 8x + 10:
f'(x) = 4x^3 - 12x + 8
Setting f'(x) equal to zero and solving for x:
4x^3 - 12x + 8 = 0
To solve this equation for x, we can factor out a common factor of 4:
4(x^3 - 3x + 2) = 0
Now, we can solve the cubic equation x^3 - 3x + 2 = 0. However, this equation does not have any rational roots, so we need to use other methods, such as synthetic division or a graphing calculator, to find its roots. By using a graphing calculator or any other method, we find that the roots are x ≈ -1.62 and x ≈ -0.37.
Now, we have two critical points: x = -1.62 and x = -0.37. To determine their nature, we can analyze the concavity of the function using the second derivative test.
Taking the derivative of f'(x) = 4x^3 - 12x + 8:
f''(x) = 12x^2 - 12
Plugging the critical points into f''(x):
f''(-1.62) = 12(-1.62)^2 - 12 ≈ -37.166
f''(-0.37) = 12(-0.37)^2 - 12 ≈ -12.372
Since f''(-1.62) is negative, the critical point x = -1.62 corresponds to a local maximum.
Since f''(-0.37) is also negative, the critical point x = -0.37 corresponds to a local maximum.
Therefore, the critical points of f(x) = x^4 - 6x^2 + 8x + 10 are x = -1.62 (local maximum) and x = -0.37 (local maximum).
To find the critical points of the function and determine their nature, we need to find the derivative of the function, which will give us the slope of the tangent line at any given point. The critical points occur when the derivative is equal to zero or undefined.
Step 1: Find the derivative of the function f(x)
f'(x) = 4x^3 - 12x + 8
Step 2: Set the derivative equal to zero and solve for x
4x^3 - 12x + 8 = 0
This equation does not factor easily, so we will use numerical methods to approximate the solutions. One possible solution is x ≈ 1.26.
Step 3: Determine if the critical points are local maxima, minima, or inflection points.
To determine the nature of each critical point, we need to examine the second derivative of the function.
Step 4: Find the second derivative of the function f(x)
f''(x) = 12x^2 - 12
Step 5: Substitute the critical points found in Step 2 into the second derivative.
For x ≈ 1.26:
f''(1.26) = 12(1.26)^2 - 12 ≈ 13.92
Since the second derivative is positive at x ≈ 1.26, the function f(x) has a local minimum at this point.
Therefore, the critical point of the function f(x) = x^4 - 6x^2 + 8x + 10 is (approximately) (1.26, f(1.26)) and it is a local minimum.