What are the local extrema, if any, of #f(x)= (x^2 + 6x-3)*e^x + 8x –8#?
This function has no local extrema.
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To find the local extrema of ( f(x) = (x^2 + 6x - 3)e^x + 8x - 8 ), we need to find the critical points by setting the derivative equal to zero and then test these points for local extrema using the second derivative test.
First, find the first derivative: [ f'(x) = (2x + 6)e^x + (x^2 + 6x - 3)e^x + 8 ]
Next, set ( f'(x) = 0 ) and solve for ( x ): [ (2x + 6)e^x + (x^2 + 6x - 3)e^x + 8 = 0 ] [ e^x(2x + 6 + x^2 + 6x - 3) + 8 = 0 ] [ e^x(x^2 + 8x + 3) + 8 = 0 ] [ e^x(x^2 + 8x + 11) = -8 ]
The solution to this equation is not straightforward algebraically, so numerical methods or approximation techniques may be needed. Once the critical points are found, you can use the second derivative test to determine whether they are local maxima or minima.
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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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