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A stationary point is where the derivative of a function is equal to zero. Let's find it for f(x)= x(x+1)^.5. Using the product rule and power rule, we compute f'(x)=x(.5)(x+1)^-.5+(x+1)^.5=[x(.5)+(x+1)]/[(x+1)^.5]. We now set this equal to zero. We can now ignore the denominator because as long as the numerator equals zero and the denominator does not equal zero, the whole thing equals zero. So 0=x(.5)+(x+1)=.5x+x+1=1.5x+1-->1.5x=-1-->x=-1/1.5=-2/3. Note that x=-2/3 does not make the denominator zero, so we have found the only stationary point. Note that f(-2/3)=(-2/3)(-2/3+1)^.5=(-2/3)(1/3)^.5≈-0.3849. That's it.

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6y ago
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6y ago

y = x*sqrt(x + 1)dy/dx = sqrt(x + 1) + x*1/2*1/sqrt(x + 1)

dy/dx = 0 => sqrt(x + 1) + x*1/2*1/sqrt(x + 1) = 0

sqrt(x + 1) = - x*1/2*1/sqrt(x + 1)

Multiplying both sides by sqrt(x + 1) [it can be shown that sqrt(x + 1) is not 0 at the turning point.]

(x + 1) = -x/2 or 2x + 2 = -x => 3x = -2 or x = -3/2.

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Q: How do you find the coordinates of the stationary point on the graph of y equals x root(x plus 1)?
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