Answer:
x1 = -0.2529 (Maximum point)
x2 = 0.3042 (Minimum point)
Step-by-step explanation:
Critical point of any polynomial may be by calculating f'(x) = 0. To further classify each critical point as minimum or maximum, we substitute the critical points in second derivative of the function. If the answer of f"(x) is positive, the respective critical point is minimum, if the value is negative then the critical point is maximum.
Step 1: Differentiate the function f(x)
[tex]f(x) = 13x^{3} - x^{2} - 3x +10\\f'(x) = 39x^{2} - 2x - 3[/tex]
Step 2: f'(x) = 0
[tex]f'(x) = 39x^{2} -2x -3 \\39x^{2} -2x -3 = 0[/tex]
Step 3: Solving the quadratic equation using the formula:
[tex]39x^{2} -2x -3 = 0\\\\x= \frac{-(-2)\ \pm \ \sqrt{(-2)^2 - 4 \times 39 \times (-3)}}{2 \times 39} \\\\x = \frac{2\ \pm \sqrt{4+468}}{78}\\\\x = \frac{2\ \pm \sqrt{472}}{78}\\\\x= \frac{2 \pm 2 \sqrt {118}}{78}\\[/tex]
[tex]x_{1} = \frac{2 + 2 \sqrt{118}}{78} \ \ \ \ \ x_{2} = \frac{2 - 2 \sqrt{118}}{78}\\\\x_{1} = -0.2529 \ \ \ \ \ x_{2} = 0.3042[/tex]
These are the two critical points of the function.
Step 4: Calculating second derivative of the function
[tex]f(x) = 13x^{3} - x^{2} - 3x +10\\f'(x) = 39x^{2} - 2x - 3\\\\f''(x) = 78x - 2[/tex]
Step 5: Substituting values of x1 and x2 in f''(x)
[tex]f''(x) = 78x - 2\\f''(x_{1}) = 78 \times (-0.2529) - 2\ \ \ \ \ f''(x_{2}) = 78 \times (0.3042) - 2\\f''(x_{1}) = -21.7262 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f''(x_{2}) = 21. 7276\\[/tex]
As stated earlier, if the value of f''(x) is negative, the point is maximum and vice versa.
[tex]x_{1}[/tex] ⇒ maximum point
[tex]x_{2}[/tex] ⇒ minimum point
