Answer :
Answer:
The following classification is found:
[tex](0.2, 6)[/tex] - Absolute minimum
[tex](-0.2, 10)[/tex] - Absolute maximum
Step-by-step explanation:
Let be [tex]f(x) = 125\cdot x^{3}-15\cdot x + 8[/tex], we need to find first and second derivatives of this expression at first:
First derivative
[tex]f'(x) = 375\cdot x^{2}-15[/tex] (Eq. 1)
Second derivative
[tex]f''(x) = 750\cdot x[/tex] (Eq. 2)
Critical points are points that equals first derivative to zero and that may be maxima or minima. That is:
[tex]375\cdot x^{2} -15 = 0[/tex]
[tex]x = \pm \sqrt{\frac{15}{375} }[/tex]
Which leads to the following critical points:
[tex]x_{1}\approx 0.2[/tex] and [tex]x_{2} \approx -0.2[/tex]
Now we evaluate each result in second derivative expression:
[tex]f''(x_{1}) = 750\cdot (0.2)[/tex]
[tex]f''(x_{1})=150[/tex] (Absolute minimum)
[tex]f''(x_{2})= 750\cdot (-0.2)[/tex]
[tex]f''(x_{2}) = -150[/tex] (Absolute maximum)
Lastly we evaluate the function at each critical point:
[tex]f(x_{1})= 125\cdot (0.2)^{3}-15\cdot (0.2)+8[/tex]
[tex]f(x_{1})= 6[/tex]
[tex]f(x_{2})= 125\cdot (-0.2)^{3}-15\cdot (-0.2)+8[/tex]
[tex]f(x_{2}) = 10[/tex]
And the following classification is found:
[tex](0.2, 6)[/tex] - Absolute minimum
[tex](-0.2, 10)[/tex] - Absolute maximum