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3 years ago

Please Help - Point x^3-4x<-5x^2+20 Solve in interval notation and list the zeros

Mathematics

1 answer:

Naddika [18.5K]3 years ago

4 0

hi my name is antonette Jane S pascua please help me my question thank you for your answer

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Simplify -2.5·4 thanks hhhhhhhhh

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Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu

kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

$f(x) = {x}^{5} - {x}^{3} + 6$

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

$f'(x) = 5 {x}^{4} - 3 {x}^{2}$

At turning point f'(x)=0.

$5 {x}^{4} - 3 {x}^{2} = 0$

This implies that,

${x}^{2} (5 {x}^{2} - 3) = 0$

${x}^{2} = 0 \: or \: 5 {x}^{2} - 3 = 0$

$x = - \frac{ \sqrt{15} }{5} \: or \: x = 0 \: \: or \: x =\frac{ \sqrt{15} }{5}$

We plug this values into the original function to obtain the y-values of the turning points

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} +750)) \:and \: (0, - 6) \: and\: ( \frac{ \sqrt{15} }{5} , \frac{1}{125} ( - 6 \sqrt{15} +750))$

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

$f''(x) = 20 {x}^{3} - 6x$

$f''( - \frac{ \sqrt{15} }{5} ) \: < \: 0$

Hence

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} + 750))$

is a maximum point.

$f''( \frac{ \sqrt{15} }{5} ) \: > \: 0$

Hence

$( \frac{ \sqrt{15} }{5} , \frac{1}{125} (- 6 \sqrt{15} + 750))$

is a minimum point.

$f''(0) \: =\: 0$

Hence (0,-6) is a point of inflexion

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3 years ago

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Answer:

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Step-by-step explanation:

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