Answer
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Hint: Intervals of monotonicity means the intervals in which the function is increasing or decreasing for which the condition is –
$ f'(x) > 0 $ for Increasing (I)
$ f'(x) < 0 $ for Decreasing (D)
Use chain rule to differentiate and calculate the roots of derivatives. Hence, insert the values to find where the function is increasing or decreasing.
Formula Used:
$ f'(x) > 0 $ for Increasing (I)
$ f'(x) < 0 $ for Decreasing (D)
Complete step-by-step answer:
We have the function \[f(x) = 2.{e^{{x^2} - 4x}}\].
Differentiating above function and applying chain rule, we get,
$ f'(x) = 2(2x - 4).{e^{{x^2} - 4x}} $
For Increasing: $ f'(x) > 0 $ i.e.
$ f'(x) = 2(2x - 4).{e^{{x^2} - 4x}} > 0 $
$ \because $ $ {e^{{x^2} - 4x}} > 0 $ for all values of x . So for $ f'(x) > 0 $ we have,
$ 2(2x - 4) > 0 $
$ \Rightarrow 4(x - 2) > 0 $
$ \Rightarrow (x - 2) > 0 $ (Since, 4 is constant)
$ \Rightarrow x > 2 $
$ \therefore $ For all values of $ x > 2 $ , $ f(x) $ is increasing
i.e. $ f(x) $ is increasing in the interval $ (2,\infty ) $
For Decreasing: $ f'(x) < 0 $
$ f'(x) = 2(2x - 4).{e^{{x^2} - 4x}} < 0 $
$ \because $ $ {e^{{x^2} - 4x}} > 0 $ for all values of x . So for $ f'(x) < 0 $ we have,
$ 2(2x - 4) < 0 $
$ \Rightarrow 4(x - 2) < 0 $
$ \Rightarrow (x - 2) < 0 $ (Since, 4 is constant)
$ \Rightarrow x < 2 $
$ \therefore $ For all values of $ x < 2 $ , $ f(x) $ is decreasing
i.e. $ f(x) $ is decreasing in the interval $ ( - \infty ,2) $
So, $ f(x) $ is increasing in the interval $ (2,\infty ) $ and decreasing in the interval $ ( - \infty ,2) $
So, the correct answer is “Option B”.
Note: The monotonicity of the function tells us whether the function is increasing or decreasing. Monotonic functions are linear, quadratic, and logarithmic. When a function is increasing on its entire domain or decreasing on its entire domain, we can say that the function is strictly monotonic. If we are given to calculate monotonicity of the function, firstly differentiate the given function and then proceed with further solution and do not forget to mention all the formulas you used.
$ f'(x) > 0 $ for Increasing (I)
$ f'(x) < 0 $ for Decreasing (D)
Use chain rule to differentiate and calculate the roots of derivatives. Hence, insert the values to find where the function is increasing or decreasing.
Formula Used:
$ f'(x) > 0 $ for Increasing (I)
$ f'(x) < 0 $ for Decreasing (D)
Complete step-by-step answer:
We have the function \[f(x) = 2.{e^{{x^2} - 4x}}\].
Differentiating above function and applying chain rule, we get,
$ f'(x) = 2(2x - 4).{e^{{x^2} - 4x}} $
For Increasing: $ f'(x) > 0 $ i.e.
$ f'(x) = 2(2x - 4).{e^{{x^2} - 4x}} > 0 $
$ \because $ $ {e^{{x^2} - 4x}} > 0 $ for all values of x . So for $ f'(x) > 0 $ we have,
$ 2(2x - 4) > 0 $
$ \Rightarrow 4(x - 2) > 0 $
$ \Rightarrow (x - 2) > 0 $ (Since, 4 is constant)
$ \Rightarrow x > 2 $
$ \therefore $ For all values of $ x > 2 $ , $ f(x) $ is increasing
i.e. $ f(x) $ is increasing in the interval $ (2,\infty ) $
For Decreasing: $ f'(x) < 0 $
$ f'(x) = 2(2x - 4).{e^{{x^2} - 4x}} < 0 $
$ \because $ $ {e^{{x^2} - 4x}} > 0 $ for all values of x . So for $ f'(x) < 0 $ we have,
$ 2(2x - 4) < 0 $
$ \Rightarrow 4(x - 2) < 0 $
$ \Rightarrow (x - 2) < 0 $ (Since, 4 is constant)
$ \Rightarrow x < 2 $
$ \therefore $ For all values of $ x < 2 $ , $ f(x) $ is decreasing
i.e. $ f(x) $ is decreasing in the interval $ ( - \infty ,2) $
So, $ f(x) $ is increasing in the interval $ (2,\infty ) $ and decreasing in the interval $ ( - \infty ,2) $
So, the correct answer is “Option B”.
Note: The monotonicity of the function tells us whether the function is increasing or decreasing. Monotonic functions are linear, quadratic, and logarithmic. When a function is increasing on its entire domain or decreasing on its entire domain, we can say that the function is strictly monotonic. If we are given to calculate monotonicity of the function, firstly differentiate the given function and then proceed with further solution and do not forget to mention all the formulas you used.
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