Your answers seem to be on the right track. These online homework apps can be picky about the answer they accept, though.
Given
, we have derivative
![f'(x) = 4x^3 - 4x = 4x (x^2 - 1) = 4x (x - 1) (x + 1)](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%204x%5E3%20-%204x%20%3D%204x%20%28x%5E2%20-%201%29%20%3D%204x%20%28x%20-%201%29%20%28x%20%2B%201%29)
with critical points when
; this happens when
or
.
We also have <em>second</em> derivative
![f''(x) = 12x^2 - 4 = 12 \left(x^2-\dfrac13\right) = 12 \left(x - \dfrac1{\sqrt3}\right) \left(x + \dfrac1{\sqrt3}\right)](https://tex.z-dn.net/?f=f%27%27%28x%29%20%3D%2012x%5E2%20-%204%20%3D%2012%20%5Cleft%28x%5E2-%5Cdfrac13%5Cright%29%20%3D%2012%20%5Cleft%28x%20-%20%5Cdfrac1%7B%5Csqrt3%7D%5Cright%29%20%5Cleft%28x%20%2B%20%5Cdfrac1%7B%5Csqrt3%7D%5Cright%29)
with (possible) inflection points when
.
Intercept
If "intercept" specifically means
-intercept, what you have is correct. Setting
gives
, so the intercept is the point (0, 3).
They could also be expecting the
-intercepts, in which case we set
and solve for
. However, we have
![x^4 - 2x^2 + 3 = \left(x^2 - 1\right)^2 + 2](https://tex.z-dn.net/?f=x%5E4%20-%202x%5E2%20%2B%203%20%3D%20%5Cleft%28x%5E2%20-%201%5Cright%29%5E2%20%2B%202)
and
![x^2-1\ge-1 \implies (x^2-1)^2 \ge (-1)^2 = 1 \implies x^4-2x^2+3 \ge 2](https://tex.z-dn.net/?f=x%5E2-1%5Cge-1%20%5Cimplies%20%28x%5E2-1%29%5E2%20%5Cge%20%28-1%29%5E2%20%3D%201%20%5Cimplies%20x%5E4-2x%5E2%2B3%20%5Cge%202)
so there are no
-intercepts to worry about.
Relative minima/maxima
Check the sign of the <em>second</em> derivative at each <em>critical</em> point.
![f''(-1) = 8 > 0 \implies \text{rel. min.}](https://tex.z-dn.net/?f=f%27%27%28-1%29%20%3D%208%20%3E%200%20%5Cimplies%20%5Ctext%7Brel.%20min.%7D)
![f''(0) = -4 < 0 \implies \text{rel. max.}](https://tex.z-dn.net/?f=f%27%27%280%29%20%3D%20-4%20%3C%200%20%5Cimplies%20%5Ctext%7Brel.%20max.%7D)
![f''(1) = 8 > 0 \implies \text{rel. min.}](https://tex.z-dn.net/?f=f%27%27%281%29%20%3D%208%20%3E%200%20%5Cimplies%20%5Ctext%7Brel.%20min.%7D)
So we have two relative minima at the points (-1, 2) and (1, 2), and a relative maximum at (0, 3).
Inflection points
Simply evaluate
at each of the candidate inflection points found earlier.
![f\left(-\dfrac{\sqrt3}3\right) = \dfrac{22}9](https://tex.z-dn.net/?f=f%5Cleft%28-%5Cdfrac%7B%5Csqrt3%7D3%5Cright%29%20%3D%20%5Cdfrac%7B22%7D9)
![f\left(\dfrac{\sqrt3}3\right) = \dfrac{22}9](https://tex.z-dn.net/?f=f%5Cleft%28%5Cdfrac%7B%5Csqrt3%7D3%5Cright%29%20%3D%20%5Cdfrac%7B22%7D9)