if y= 147x^4+28x^3-12x^2+10 on the closed interval [-20,20], use calculus and sign analysis to find the exact values for the x-coordinates of all ...if y= 147x^4+28x^3-12x^2+10 on the closed...

if y= 147x^4+28x^3-12x^2+10 on the closed interval [-20,20], use calculus and sign analysis to find the exact values for the x-coordinates of all ...

if y= 147x^4+28x^3-12x^2+10 on the closed interval [-20,20], use calculus and sign analysis to find the exact values for the x-coordinates of all absolute maxima, absolute minima,local maxima,local minima.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `f(x)=147x^4+28x^3-12x^2+10` on [-20,20]:

The extrema must occur at critical points -- since this is a polynomial the critical points will be where f'(x)=0. To check for absolute maximum and minimum we also check the endpoints.

`f'(x)=588x^3+84x^2-24x`

`=12x(49x^2+7x-2)`

`=12x(7x-1)(7x+2)`

Letting f'(x)=0 we get the following critical points:

`x=0,1/7,-2/7`

We evaluzte f(x) at the critical points and the endpoints of the interval:

`f(-20)=23291210`

`f(-2/7)~~9.3469`

`f(0)=10`

`f(1/7)~~9.89796`

`f(20)=23739210`

So the function on this interval has an absolute maximum at x=20.

Checking convenient values we find that on the interval:

`[-20,-2/7):f'(x)<0` so the function is decreasing on this interval

`(-2/7,0):f'(x)>0` so the function is increasing on this interval.

At `x=-2/7` there is an absolute minimum on the interval.

`(0,1/7):f'(x)<0` so the function is decreasing on this interval.

There is a local maximum at x=0.

`(1/7,20]:f'(x)>0` so the function is increasing on this interval.

There is a local minimum at `x=1/7` .

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