The function is obviously defined only for `v gt 0` and is continuously differentiable on this interval. When `v` approaches zero the function tends to `+oo,` when v tends to `+oo,` the function also tends to `+oo.`

Thus it must have at least one local (and global) minimum and it is reached at the point(s) where `P'(x) = 0.` Let's solve this equation:

`P'(x) = -17 v^-2 + 3*10^-3*v^2 = 0.`

This is equivalent to `17 v^-2 = 3*10^-3*v^2,` or `v^4 = 17/3*10^3.`

The only solution is `v = root(4)(17/3*10^3) approx` **8.7** (m/s). This is the answer.

Take the derivative of `P(v)` .

`P'(v)=-17v^-2+3*10^-3v^2`

Set `P'(v)` equal to zero and solve for the critical values.

`-17v^-2+3*10^-3v^2=0`

`-17+3*10^-3v^4=0`

`3*10^-3v^4=17`

`v^4=17/3*10^3`

`v^4=17/3*10^3`

`v=(17/3*10^3)^(1/4)~~8.676 m/s`

We can check graphically that this is indeed a minimum.

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