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Find the local maximum, the local minimu...

Find the local maximum, the local minimum values, if any, using first or second derivative test of the function f defined by
`f(x)=1+1/x+1/x^2`

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To find the local maximum and minimum values of the function \( f(x) = 1 + \frac{1}{x} + \frac{1}{x^2} \), we will use the first and second derivative tests. ### Step 1: Find the first derivative \( f'(x) \) The first step is to differentiate the function \( f(x) \). \[ f(x) = 1 + \frac{1}{x} + \frac{1}{x^2} ...
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MAHAVEER PUBLICATION-DIFFERENTIATION OR DERIVATIVE OF A FUNCTION-QUESTION BANK
  1. Without using derivative, find the maximum value and the minimum value...

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  2. Find the local maximum, the local minimum values, if any, using first ...

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  3. Find the local maximum, the local minimum values, if any, using first ...

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  4. Find the values of x, if any, for which f(x) has local maximum and loc...

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  5. Find the values of x, if any, for which f(x) has local maximum and loc...

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  6. Find the values of x, if any, for which f(x) has local maximum and loc...

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  7. Find the values of x, if any, for which f(x) has local maximum and loc...

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  8. Find the local maximum, local minimum, absolute maximum and absolute m...

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  9. Find the local maximum, local minimum, absolute maximum and absolute m...

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  10. Show that f(x)=3x-x^3 has a local maximum at x=1.

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  11. Show that f(x)=x^2+(250)/x has a local minimum at x=5.

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  12. Show that f(x)=1+x+x^2+x^3 has no local extremum.

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  13. Show that f(x)=x^3-3x^2+15x+2 has no local extremum.

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  14. Find the extreme values of the function y=2x^3-9x^2+12x+5

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  15. Find the maximum and minimum values of the function y=4x^3-3x^2-6x+1

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  16. Find two positive numbers x and y such that x+y=28 and xy is maximum.

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  17. Find two positive numbers x and y such that x+y=64 and x^3+y^3 is min...

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  18. Find two positive numbers x and y such that x+y=400 and xy^3 is maxim...

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  19. Find two positive numbers x and y such that xy=36 and x+y is minimum.

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  20. Find the point on the curve y^2=4x which is nearest to the point...

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