# Math Help - Help for a function please

1. ## Help for a function please

I don't have the slightest idea how to find the turning points of this function; f(x) = 2/3X^3 - 1/2X^2 + 4
I'd appreciate it a lot

2. Originally Posted by FionaG
I don't have the slightest idea how to find the turning points of this function; f(x) = 2/3X^3 - 1/2X^2 + 4
I'd appreciate it a lot
You have to differentiate that function and have it equal 0 .

$f(x)=\frac{2}{3}x^3-\frac{1}{2}x^2+4$

$\frac{dy}{dx}=2x^2-x$

$2x^2-x=0$ , find the values of x and sub them into the original equation to get the y's .

3. Thanks, but could anyone tell me the result? I just don't understand this one

4. $x(2x-1)=0\rightarrow x=0\text{ or }\frac 12\Rightarrow y=4\text{ or }\frac 23\cdot\left(\frac 12\right)^3-\frac 12\cdot \left(\frac 12\right)^3+4\approx 3.958$. So the points are: $(0|4)$, $\left(\frac 14|\approx 3.958\right)$.