Square Root Problem
- Thread starter Jason76
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D
Deleted member 4993
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\(\displaystyle f'(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)Book says this is wrong.
Your algebra in these last two steps are correct.
May be the equation you started with - is incorrect.
\(\displaystyle f'(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
That what you have written is right!!!! \(\displaystyle \sqrt{x}(x - 14)\) is equal to \(\displaystyle x^{3/2} - 14x^{1/2}\) !!
What is the answer of your book??
D
Deleted member 4993
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\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (3/2)*x^{3/2-1} - 14*(1/2)*x^{(1/2-1)}\)
\(\displaystyle f'(x) = (3/2)*x^{1/2} - 7*x^{(-1/2)}\)
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
Last edited:
D
Deleted member 4993
Guest
Online homework said this was correct
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
- What do you mean? It's only rewriting the square root.
D
Deleted member 4993
Guest
Online homework said this was correct
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
f'(x) = \(\displaystyle \frac{df(x)}{dx} = \) \(\displaystyle x^{1/2}(x - 14)\)......................Are you claiming \(\displaystyle \frac{df(x)}{dx} = \) f(x) ??
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
- What do you mean? It's only rewriting the square root.
I mean - that you are being careless!!
Online homework said this was correct
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f'(x) = x^{1/2}(x - 14)\).....NO, this is f(x), not f'(x)
\(\displaystyle f'(x) = x^{3/2} - 14x^{1/2}\).....NO, this is f(x), not f'(x)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
..................FINALLY you have differentiated!
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
- What do you mean? It's only rewriting the square root.
When you are careless in the intermediate steps, we wonder how you get back on track.Subhotosh Kahn said:
Ok, fixed problem. Sorry.
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f(x) = x^{1/2}(x - 14)\)
\(\displaystyle f(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer
\(\displaystyle f(x) = \sqrt{x}(x - 14)\)
\(\displaystyle f(x) = x^{1/2}(x - 14)\)
\(\displaystyle f(x) = x^{3/2} - 14x^{1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - (\dfrac{1}{2})14x^{-1/2}\)
\(\displaystyle f'(x) = (\dfrac{3}{2})x^{1/2} - 7x^{-1/2}\) - Answer