HELP ME PLEASE! AGH!

Mathematics

1 answer:

Vanyuwa [196]4 years ago

8 0

I'm a third grader!!!!!!! Not a highschooler

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

antiseptic1488 [7]

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:

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

3 0

3 years ago

Hey um i really need help i tried to do it but i couldn't do it

kenny6666 [7]

Answer:

ok i dnt know but

Step-by-step explanation:

go to https://mathsolver.microsoft.com/en/solve-problem/2%20%60left(%20y-3%20%60cdot%20%204%20%20%60right) if it dnt work try a calculator

7 0

3 years ago

What is the equation of the line

Delvig [45]