The function f has first derivative given by f'(x)=√x/1+x+x^3. What is the x-coordinate of the inflection point of the graph of f?

A)1.008
B)0.473
C)0
D)-0.278
E)no point of inflection

Question

The function f has first derivative given by f'(x)=√x/1+x+x^3. What is the x-coordinate of the inflection point of the graph of f?

A)1.008
B)0.473
C)0
D)-0.278
E)no point of inflection

polaris_s0i · Answer

is this you're equation?
$$f^\prime(x) = \frac{\sqrt{x}}{1} + x +x^3$$

anonymous · Answer

to get the point of inflection, calculate the second derivative of the function and solve for it zeros, the value(s) of x that will make the function zero is the inflection point

anonymous · Answer

I got
f''(x)=[1/2(√x)*(1+x+x^3)-(1+3x^2)*(√x)]/(1+x+x^3)
but how do i continue this after setting it to 0?

polaris_s0i · Answer

oh this is you're function:
$$f'(x)=\frac{\sqrt{x}}{1+x+x^3}$$

polaris_s0i · Answer

you need to square the bottom:
$$f''(x)=\frac{\frac{1}{2}\sqrt{x}(1+x+x^3)-(1+3x^2)\sqrt{x}}{(1+x+x^3)^2}$$
trying to figure out how to factor this beast.

polaris_s0i · Answer

sorry its this actually.  we both didn't decrement g to -1/2
$$f''(x)=\frac{\frac{1}{2\sqrt{x}}(1+x+x^3)-(1+3x^2)\sqrt{x}}{(1+x+x^3)^2}$$