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Prove that the function f given by f(x)=...

Prove that the function f given by `f(x)=x^2-x+1`is neither strictly increasing nor strictly decreasing on `( 1, 1)`.

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To prove that the function \( f(x) = x^2 - x + 1 \) is neither strictly increasing nor strictly decreasing on the interval \( (-1, 1) \), we will follow these steps: ### Step 1: Find the derivative of the function To analyze the behavior of the function, we first need to find its derivative. \[ f'(x) = \frac{d}{dx}(x^2 - x + 1) = 2x - 1 \] ...
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NCERT-APPLICATION OF DERIVATIVES-EXERCISE 6.2
  1. Prove that y=(4sintheta)/((2+costheta) )-thetais an increasing functio...

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  2. Find the values of x for which y=[x(x-2)]^2is an increasing function

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  3. Show that the function given by f (x) = 3x + 17is strictly increasing...

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  4. Show that the function given by f (x) = sinx is (a) strictly increas...

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  5. Show that the function given by f(x)=e^(2x) is strictly increasing on...

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  6. Find the intervals in which the function f given by f(x)=2x^3-3x^2-36 ...

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  7. Find the intervals in which the function f given by f(x)=2x^2-3xis (a)...

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  8. Show that y=log(1+x)-(2x)/(2+x), x gt-1, is an increasing function of ...

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  9. Find the intervals in which the following functions are strictly incre...

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  10. Prove that the function given by f(x)=x^3-3x^2+3x-100is increasing in...

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  11. The interval in which y=x^2e^(-x)is increasing is(A) (-oo, oo) (B) ( ...

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  12. Which of the following functions are strictly decreasing on (0,pi/2) ...

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  13. On which of the following intervals is the function f given by f(x)=x...

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  14. Prove that the logarithmic function is strictly increasing on (0,oo).

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  15. Prove that the function f given by f(x)=x^2-x+1is neither strictly in...

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  16. Prove that the function f given by f(x) = log sin xf(x) = log sin xis ...

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  17. Prove that the function f given by f (x) = log cos x is strictly decre...

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  18. Find the least value of a such that the function f given by f(x)=x^2+...

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  19. Let I be an interval disjointed from [-1,\ 1] . Prove that the functio...

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