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Alexxandr [17]
3 years ago
12

Use CALCULUS to find coordinates of the turning point on C.

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
3 0
Oi 

Just calculating the differential. Step by step:

y=12 \sqrt{x} -x \frac{3}{2}-10 \\  \\ y=12x^{ \frac{1}{2} } -x \frac{3}{2}-10 \ \ \ \boxed{transform \  \sqrt{x} =x^{ \frac{1}{2} }}  \\  \\  Differentiating \\  \\ y'=12. \frac{1}{2}.x^{ \frac{1}{2}-1 } -1.x^{1-1}. \frac{3}{2}-0 \\  \\  y'=6.x^{ -\frac{1}{2} } -x^{0}. \frac{3}{2} \\  \\  y'=6. \frac{1}{x^{ \frac{1}{2} }}  -1. \frac{3}{2}  \\  \\  \boxed{y'=\frac{6}{ \sqrt{x} }  -\frac{3}{2}}

If you want to know where the tangent has no slope:

\frac{6}{ \sqrt{x} } - \frac{3}{2} =0 \\  \\  \frac{6}{ \sqrt{x} } = \frac{3}{2}  \\  \\ 3 \sqrt{x} =12 \\  \\  \sqrt{x} = \frac{12}{3}  \\  \\  \sqrt{x} =4 \\  \\   (\sqrt{x})^2 =4^2 \\  \\ x=16
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