If a curve has the gradient, g'(x) = k / (x^2) - [square-root of (x)]
where k is a constant and a stationary point at (4 , -2), then the value of k is?
I think u have to integrate it or something.
The answers is 32? but can someone show me how they get 32 in a fully worked step-by-step solution.
2007-08-21 21:02:14 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Thanks all :D
2007-08-21 21:21:22 · update #1
g'(x) = k/x^2 - √x
If there is a stationary point at x=4, then g'(4) = 0.
So k/16 - √4 = 0
=> k/16 = 2
=> k = 32.
2007-08-21 21:09:47 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 1⤋
Is the stationary point refering to the gradient or the original curve? If the original curve then:
Integrate g'(x) = g(x) + C = -k/x - (x^1.5)/1.5 + C
now plug in (4, -2)
g(x)= -2 = -k/4 - (4*2)/1.5
-k/4 = -2+ 8/1.5
k/4 = 2 - 16 / 3 = -10/3
k = -40/3
If it refers to the gradient then see other answers.
2007-08-22 04:17:40 · answer #2 · answered by snapmedown 2 · 0⤊ 2⤋
Plugging 4 in g'(x) = kx^{- 2} - x^{1/2} gives
0 = g'(4) = k/16 - 2 and therefore k = 2.16 = 32. (No need to compute g!!!)
2007-08-22 04:14:51 · answer #3 · answered by polizei 2 · 0⤊ 1⤋