# How do you find the critical points of #h'(x)=x^2+8x-9#?

Critical points of

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To find the critical points of the function ( h'(x) = x^2 + 8x - 9 ), you need to first find its derivative, then set it equal to zero and solve for ( x ).

- Find the derivative of ( h'(x) ) which is ( h''(x) = 2x + 8 ).
- Set the derivative equal to zero: ( 2x + 8 = 0 ).
- Solve for ( x ): ( 2x = -8 ), ( x = -4 ).

So, the critical point of ( h'(x) ) is ( x = -4 ).

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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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