How do you find all extrema for #f(x) = 2x + ln x#?

Answer 1
Find all extrema for #f(x) = 2x + ln x#
#f'(x) = 2+1/x = (2x+1)/x#
#f'(x) =0# at #x=-1/2# and is not defined at #x=0#, but neither #-1/2# nor #0# is in the domain of #f#, so there are no critical numbers. Therefore, there are no extrema.

graph{y=2x+lnx [-18, 18.02, -8.99, 9.04]}

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Answer 2

To find all extrema for ( f(x) = 2x + \ln(x) ), you first need to find the critical points by setting the derivative equal to zero and solving for ( x ). Then, you can classify these critical points as either maxima or minima using the second derivative test.

  1. Find the derivative of ( f(x) ): ( f'(x) = 2 + \frac{1}{x} ).
  2. Set ( f'(x) ) equal to zero and solve for ( x ): [ 2 + \frac{1}{x} = 0 ] [ \frac{1}{x} = -2 ] [ x = -\frac{1}{2} ]
  3. Calculate the second derivative ( f''(x) ): [ f''(x) = -\frac{1}{x^2} ]
  4. Evaluate ( f''(-\frac{1}{2}) ) to determine the concavity. [ f''(-\frac{1}{2}) = -\frac{1}{(-\frac{1}{2})^2} = -4 ]
  5. Since the second derivative is negative at ( x = -\frac{1}{2} ), this critical point is a maximum.

Therefore, the only extremum for ( f(x) = 2x + \ln(x) ) is a maximum at ( x = -\frac{1}{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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