# Is #f(x)=e^(3-3x)+x/ln2x# concave or convex at #x=1#?

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To determine whether ( f(x) = e^{3-3x} + \frac{x}{\ln(2x)} ) is concave or convex at ( x = 1 ), we need to analyze the second derivative of ( f(x) ) at ( x = 1 ).

The second derivative test states that if the second derivative of a function is positive at a critical point, then the function is concave up at that point; if the second derivative is negative, then the function is concave down at that point.

First, we find the first derivative of ( f(x) ) using the sum rule and the chain rule:

( f'(x) = -3e^{3-3x} + \frac{1}{\ln(2x)} - \frac{x}{(2x)\ln^2(2x)} )

Next, we find the second derivative of ( f(x) ):

( f''(x) = 9e^{3-3x} + \frac{-1}{x(\ln(2x))^2} - \frac{1}{(2x)\ln(2x)} + \frac{1}{(2x^2)\ln^2(2x)} )

Now, we evaluate ( f''(1) ) to determine concavity at ( x = 1 ):

( f''(1) = 9e^{3-3(1)} - \frac{1}{1(\ln(2))^2} - \frac{1}{2(1)\ln(2)} + \frac{1}{2(1)^2\ln^2(2)} )

( f''(1) = 9e^0 - \frac{1}{(\ln(2))^2} - \frac{1}{2\ln(2)} + \frac{1}{2\ln^2(2)} )

( f''(1) = 9 - \frac{1}{(\ln(2))^2} - \frac{1}{2\ln(2)} + \frac{1}{2\ln^2(2)} )

Since ( f''(1) ) is positive, ( f(x) ) is concave up at ( x = 1 ).

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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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