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...
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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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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