What are the critical values, if any, of #f(x)= 5x + 6x ln x^2#?
# x = e^(-17/12) ~~ 0.659241 \ \ (6 \ dp)#
We have:
Which, using the properties of logarithms, we can write as:
At a critical point, we requite that the first derivative vanishes, thus we require that:
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To find the critical values of ( f(x) = 5x + 6x \ln(x^2) ), we need to take the derivative of ( f(x) ), set it equal to zero, and solve for ( x ).
The derivative of ( f(x) ) is ( f'(x) = 5 + 6(1 + \ln(x^2)) \cdot \frac{d}{dx}(x^2) ).
Simplify the derivative: ( f'(x) = 5 + 6(1 + \ln(x^2)) \cdot 2x ).
Now, set ( f'(x) ) equal to zero and solve for ( x ): ( 5 + 12x(1 + \ln(x^2)) = 0 ).
( 5 + 12x + 12x \ln(x^2) = 0 ).
Using properties of logarithms, ( \ln(x^2) = 2\ln(x) ):
( 5 + 12x + 24x \ln(x) = 0 ).
There's no direct algebraic way to solve this equation. Critical values can be determined numerically or graphically.
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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.
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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