What is the slope of the line normal to the tangent line of #f(x) = cosx-cos^2x +1 # at # x= (17pi)/12 #?
Slope of the normal to the tangent line
Using trigonometric formulas like half-angle formulas and sum/difference formulas
God bless.....I hope the explanation is useful.
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To find the slope of the line normal to the tangent line of (f(x) = \cos(x) - \cos^2(x) + 1) at (x = \frac{17\pi}{12}), first, find the derivative of (f(x)), then evaluate it at (x = \frac{17\pi}{12}). The slope of the tangent line is the value of the derivative at that point. The slope of the line normal to the tangent line is the negative reciprocal of this slope. So, find the derivative, evaluate it at (x = \frac{17\pi}{12}), and then take the negative reciprocal to get the slope of the line normal to the tangent line.
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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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