Question 248504
{{{f(x) = x + 3x^(2/3)}}}
The derivative of a sum is the sum of the derivatives. So
f'(x) = {{{(d/(dx))(x) + (d/(dx))3x^(2/3)}}}
I hope you know that the derivative of x is 1. For the derivative of {{{3x^(2/3)}}} we will use the bring the exponent down in front and subtract 1. This gives us:
f'(x) = {{{1 + (2/3)3x^((2/3)-1)}}}
which simplifies to:
f'(x) = {{{1 + 2x^((-1)/3)}}}<br>
For the second derivative we find the derivative of the first derviative. The derivative of a constant like 1 is zero. And we will repeat the exponent thing on the second term:
f''(x) = {{{0 + ((-1)/3)x^(((-1)/3)-1)}}}
which simplifies to:
f''(x) = {{{((-1)/3)x^((-4)/3)}}}<br>
These are the first and second derivatives. You mention something about them being zero. As a general rule derivative are not always zero.<br>
However we are often interested in the value(s) of x that make the derivatives zero. If this is what you are looking for then set the derivatives equal to zero and use Algebra to solve for x. Here's a solution for the x values that make the first derivative zero:
{{{0 = 1 + 2x^((-1)/3)}}}
Subtract 1 from each side:
{{{-1 = 2x^((-1)/3)}}}
Divide by 2:
{{{(-1)/2 = x^((-1)/3)}}}
Raise each side to the -3 power:
{{{((-1)/2)^(-3) = (x^((-1)/3))^(-3)}}}
{{{-8 = x}}}
This tells us that the slope of the tangent to f(x) is zero at x = -8.<br>
I will leave it up to you to find the x values, if any, that make the second derivative zero.