Lena1992
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Super Easy Derivative that has given me a headache!
So I have to get the derivative for f(x) = x^3/1 *(x-2)
I have to find:
a) Critical Numbers
b) Intervals where it decreases/ increases
c) Local extremes
d) Concave intervals
Then the second derivative of that to get the concave intervals
I tried doing the second derivative but I seem to get a wrong answer everytime.I get f(x) = 4x-2 / 3x^2/3 for the first derivative
But I get (12x^2/3) / x^4/3 + 4^x1/3
Can somebody help? I just need to know how to get the second derivative... And no I'm not asking for someone to do my Homework.
Edit: I will post my procedure
f(x)= x^1/3(x-2)
f'(x)= (1/3x^-2/3)(x-2) + (x^1/3)(1) Here I used the product rule (f*g)' = f'*g + f*g'
f'(x)= x-2/3x^2/3 + x^1/3
f'(x)= x-2/3x^2/3 + x^1/3*(3x^2/3)/(3x^2/3) I multiply by (3x^2/3)/(3x^2/3) to make it equal
f'(x)= x-2+3x/ 3x^2/3
f'(x)= 4x-2/3x^2/3
f'(x)=4x-2/3x^2/3
f''(x)= (4)(3x^2/3) - (2^-1/3)(4x-2) / (3x^2/3)^2 Here I use the quotient rule f'(x)= f'*g - g'*f / g^2
f''(x)= 12x^2/3 - (8x^-4/3 - 4x^-1/3) / 9x^4/3
f''(x)= 12x^2/3 / 9x^4/3 - 8x^4/3 + 4x^1/3
f''(x)= 12x^2/3 / x^4/3 + 4x^1/3
Homework Statement
So I have to get the derivative for f(x) = x^3/1 *(x-2)
I have to find:
a) Critical Numbers
b) Intervals where it decreases/ increases
c) Local extremes
d) Concave intervals
Then the second derivative of that to get the concave intervals
The Attempt at a Solution
I tried doing the second derivative but I seem to get a wrong answer everytime.I get f(x) = 4x-2 / 3x^2/3 for the first derivative
But I get (12x^2/3) / x^4/3 + 4^x1/3
Can somebody help? I just need to know how to get the second derivative... And no I'm not asking for someone to do my Homework.
Edit: I will post my procedure
f(x)= x^1/3(x-2)
f'(x)= (1/3x^-2/3)(x-2) + (x^1/3)(1) Here I used the product rule (f*g)' = f'*g + f*g'
f'(x)= x-2/3x^2/3 + x^1/3
f'(x)= x-2/3x^2/3 + x^1/3*(3x^2/3)/(3x^2/3) I multiply by (3x^2/3)/(3x^2/3) to make it equal
f'(x)= x-2+3x/ 3x^2/3
f'(x)= 4x-2/3x^2/3
f'(x)=4x-2/3x^2/3
f''(x)= (4)(3x^2/3) - (2^-1/3)(4x-2) / (3x^2/3)^2 Here I use the quotient rule f'(x)= f'*g - g'*f / g^2
f''(x)= 12x^2/3 - (8x^-4/3 - 4x^-1/3) / 9x^4/3
f''(x)= 12x^2/3 / 9x^4/3 - 8x^4/3 + 4x^1/3
f''(x)= 12x^2/3 / x^4/3 + 4x^1/3
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