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Calcus help

Msjaygaab

Can someone tell me how to work out part C please

Reply 1

Zacken

Original post by Msjaygaab

Can someone tell me how to work out part C please

You've found dy/dx, haven't you? Equate it to zero and solve for x, then plug it back into the equation of the curve to find the corresponding y.

Reply 2

Msjaygaab

Yeah I got 1-32x^-5=0
How do I solve that for X?

Reply 3

Zacken

Original post by Msjaygaab

Yeah I got 1-32x^-5=0
How do I solve that for X?

Assuming that that's correct:

\displaystyle 1 - 32x^{-5} = 0 \Rightarrow 32x^{5} = 1 \Rightarrow x^5 = \frac{1}{32} \iff x^5 = \frac{1}{2^{5}} \iff x^5 = \left(\frac{1}{2}\right)^5 \iff \cdots

Reply 4

Kvothe the Arcane

Original post by Msjaygaab

Yeah I got 1-32x^-5=0
How do I solve that for X?

recall that

x^{-n}=\dfrac{1}{x^n}

and get one term on each side of the equation.

Reply 5

Mystery.

Stationary/minimum/maximum points occur when the derivative equals zero because imagine drawing a tangent at the minimum point- the gradient is 0 because it's a horizontal line.

Reply 6

TeeEm

is there any room for me?

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