# Is #f(x)=(x-16)(x-12)(x/4+3)# increasing or decreasing at #x=1#?

The gradient is negative so the function is decreasing

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To determine if ( f(x) = (x - 16)(x - 12)\left(\frac{x}{4} + 3\right) ) is increasing or decreasing at ( x = 1 ), you can use the first derivative test.

- Find the derivative of ( f(x) ).
- Substitute ( x = 1 ) into the derivative to find its value.
- If the derivative at ( x = 1 ) is positive, ( f(x) ) is increasing at that point. If it's negative, ( f(x) ) is decreasing.

[ f'(x) = (x - 16)(x - 12)\left(\frac{1}{4}\right) + (x - 16)\left(\frac{x}{4} + 3\right) + (x - 12)\left(\frac{x}{4} + 3\right) ]

[ f'(x) = \frac{(x - 16)(x - 12)}{4} + \frac{x(x - 16)}{4} + \frac{x(x - 12)}{4} + 3(x - 16) + 3(x - 12) ]

[ f'(x) = \frac{x^2 - 28x + 192 + x^2 - 16x + x^2 - 12x + 3x - 48 + 3x - 36}{4} ]

[ f'(x) = \frac{3x^2 - 53x + 108}{4} ]

Now, substitute ( x = 1 ) into ( f'(x) ) to find its value:

[ f'(1) = \frac{3(1)^2 - 53(1) + 108}{4} ] [ f'(1) = \frac{3 - 53 + 108}{4} ] [ f'(1) = \frac{58}{4} ] [ f'(1) = 14.5 ]

Since ( f'(1) > 0 ), ( f(x) ) is increasing 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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