f(x)=ln(x^4+27)
find the inflection points?
f'(x) = 4x^3/(x^4 + 27)
f''(x) = (12x^2(x^4+27) - 4x^3(4x^3))/(x^4+27)^2
= 4x^2(81-x^4)/(x^4+27)^2 = 0 for inflection points
4x^2 = 0 or 81-x^4 = 0
x = 0 or (9-x^2)(9+x)^2 = 0
x = 0 or (3+x)(3-x)(9+x)^2 = 0
x = 0 or x = -3 or x = 3
So we know the x values of our 3 points of inflection, sub into the original equation to find the matching y values.
The inflection points are where the second derivative is zero.
f'(x) = 3x^2/(x^4 +27)
f''(x) = [(x^4+27)(6x) - 12x^5]/(x^4 +27)^2
f''(x) = 0 where the denominator is zero.
i am not getting the ans i mean it says you have to find smaller and larger values.
i already know that one smaller is -3
and one larger value is 3 but the other two i am not getting :(
To find the inflection points of the function f(x)=ln(x^4+27), we need to find the second derivative of the function and then find the values of x for which the second derivative equals zero or does not exist.
First, let's find the first derivative of the function f(x) with respect to x:
f'(x) = d/dx[ln(x^4+27)]
Using the chain rule, we get:
f'(x) = (1/(x^4+27))*(d/dx)(x^4+27)
Applying the power rule for differentiation, we have:
f'(x) = (1/(x^4+27))*(4x^3)
Now, let's find the second derivative of the function f(x) by differentiating f'(x):
f''(x) = d/dx[(1/(x^4+27))*(4x^3)]
Applying the product rule, we get:
f''(x) = (1/(x^4+27)) * (d/dx)(4x^3) + (4x^3) * (d/dx)(1/(x^4+27))
Simplifying further, we have:
f''(x) = (1/(x^4+27)) * (12x^2) - (4x^3) * (1/(x^4+27))^2 * (d/dx)(x^4+27)
To find the inflection points, we need to solve the equation f''(x) = 0. So, we set f''(x) equal to zero:
(1/(x^4+27)) * (12x^2) - (4x^3) * (1/(x^4+27))^2 * (d/dx)(x^4+27) = 0
Simplifying the equation, we get:
12x^2 - 4x^3 * (1/(x^4+27))^2 * (d/dx)(x^4+27) = 0
Next, we solve this equation to find the values of x at which the second derivative equals zero. This will give us the inflection points of the function f(x)=ln(x^4+27).