The curve given by the parametric equations #x=16  t^2#, #y= t^3  1 t# is symmetric about the xaxis. At which x value is the tangent to this curve horizontal?
Slope of the tangent at a point on the cuurve, is the value of first derivative at that point.
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To find the xvalue at which the tangent to the curve is horizontal, we need to determine when the derivative of y with respect to x is zero.
Given the parametric equations (x = 16  t^2) and (y = t^3  1), we can eliminate the parameter (t) to express (y) solely in terms of (x). Then, differentiate (y) with respect to (x) and set the derivative equal to zero to find the critical points.

Eliminate the parameter (t) by solving the equation (x = 16  t^2) for (t): [t = \pm \sqrt{16  x}]

Substitute the expression for (t) into the equation for (y): [y = (\sqrt{16  x})^3  1]

Simplify the expression for (y): [y = (16  x)^{3/2}  1]

Differentiate (y) with respect to (x) to find the derivative: [\frac{dy}{dx} = \frac{3}{2}(16  x)^{1/2}]

Set the derivative equal to zero and solve for (x): [\frac{3}{2}(16  x)^{1/2} = 0] [16  x = 0] [x = 16]
So, the xvalue at which the tangent to the curve is horizontal is (x = 16).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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