Question

Find all relative extrema and classify each as a maximum or minimum. Use the second-derivative test where possible. f(x) = 125x 3 − 15x + 8

Answer 1

Answer:

The following classification is found:

$(0.2, 6)$ - Absolute minimum

$(-0.2, 10)$ - Absolute maximum

Step-by-step explanation:

Let be $f(x) = 125\cdot x^{3}-15\cdot x + 8$, we need to find first and second derivatives of this expression at first:

First derivative

$f'(x) = 375\cdot x^{2}-15$ (Eq. 1)

Second derivative

$f''(x) = 750\cdot x$ (Eq. 2)

Critical points are points that equals first derivative to zero and that may be maxima or minima. That is:

$375\cdot x^{2} -15 = 0$

$x = \pm \sqrt{\frac{15}{375} }$

Which leads to the following critical points:

$x_{1}\approx 0.2$ and $x_{2} \approx -0.2$

Now we evaluate each result in second derivative expression:

$f''(x_{1}) = 750\cdot (0.2)$

$f''(x_{1})=150$ (Absolute minimum)

$f''(x_{2})= 750\cdot (-0.2)$

$f''(x_{2}) = -150$ (Absolute maximum)

Lastly we evaluate the function at each critical point:

$f(x_{1})= 125\cdot (0.2)^{3}-15\cdot (0.2)+8$

$f(x_{1})= 6$

$f(x_{2})= 125\cdot (-0.2)^{3}-15\cdot (-0.2)+8$

$f(x_{2}) = 10$

And the following classification is found:

$(0.2, 6)$ - Absolute minimum

$(-0.2, 10)$ - Absolute maximum