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grandymaker [24]
3 years ago
5

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
Mathematics
1 answer:
MrRissso [65]3 years ago
3 0

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

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