Question

find minimum and maximum values of 10x^6-24x^5+15x^4-40x^3+108
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Answer 1

Step-by-step explanation:

this is the answer of your question

Answer 2

Concept:

In calculus, the concept of derivatives is utilised to discover maxima and minima. Any function's maxima and minima may be found using the derivative test. In most cases, first and second-order derivative tests are utilized.

Given:

The function $f(x) = 10x^6-24x^5+15x^4-40x^3+108$.

Find:

The minimum and maximum values of the function.

Solution:

$f(x) = 10x^6-24x^5+15x^4-40x^3+108$

$f'(x) = 0\\f'(x)=60x^5-120x^4+60x^3-120x^2\\f'(x)= 60x^2(x^3-2x^2+x-2)\\f'(x)= 60x^2(x-2)(x^2+1)=0\\x = 0, x=2$

$f''(x) = 300x^4-480x^3+180x^2-240x\\f''(2)=300(2)^4-480(2)^3+180(2)^2-240(2)\\f''(2)= 4800-3840+720-480\\f''(2)= 1200\\f''(2) > 0$

$f(x)$ is minimum at $x=2$.

$f''(0)=0$

$f(2)=10(2)^6-24(2)^5+15(2)^4-40(2)^3+108\\f(2) = 640-768+240-320+108\\f(2) = -100\\\\f(0) = 108$

Hence, the minimum value of the function is $-100$ and the maximum value is $108$.