C2 Differentiation question Watch

impossibleisnothing
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#1
Report Thread starter 7 years ago
#1
Here is the question:
Given that: Y = x^1.5 + 48x^-1 (x is greater than 0)
A) Find value of x and the value of y when f’(x) = 0
B) Show that the value of y which you found in a is a minimum

It seems simple, I can do the second part but i can't do a because I don't know how to solve f'(x)= 1.5x^0.5 - 48x^-2= 0.

Could someone please help me with this.
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Dizzy in my Head
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Well, use either factorisation or put it into the quadratic formula for a start. Maybe it would look easier if you put the powers as fractions rather than decimals?
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Haulward
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#3
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A)

\frac{3}{2}(x^\frac{1}{2})-48x^{-2}=0

\frac{3}{2}(x^\frac{1}{2}-32x^{-2})=0

x^ \frac{1}{2}-32x^{-2}=0

x^ \frac{1}{2}(1-32x^\frac{-5}{2})=0

1-32x\frac{-5}{2}=0 ,........... x^ \frac{1}{2}=0

\frac{32}{x^\frac{5}{2}}=1

x^5=2^{10}

x=0, x=4, y=20
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EnVogue
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For part A you need to think about what it means when dy/dx = 0. What does that mean about the curve?
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rei dos reis
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#5
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This is easy just equate f'x to o and carry out normal fraction substraction. Is x=4?
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