Question

Calculus HW , will someone please teach me how to do this problem. thanks! 10 Points

Answer 1

Your answers seem to be on the right track. These online homework apps can be picky about the answer they accept, though.

Given $f(x) = x^4 - 2x^2 + 3$, we have derivative

$f'(x) = 4x^3 - 4x = 4x (x^2 - 1) = 4x (x - 1) (x + 1)$

with critical points when $f'(x) = 0$; this happens when $x=0$ or $x=\pm1$.

We also have second derivative

$f''(x) = 12x^2 - 4 = 12 \left(x^2-\dfrac13\right) = 12 \left(x - \dfrac1{\sqrt3}\right) \left(x + \dfrac1{\sqrt3}\right)$

with (possible) inflection points when $x=\pm\frac1{\sqrt3}=\pm\frac{\sqrt3}3$.

Intercept

If "intercept" specifically means $y$-intercept, what you have is correct. Setting $x=0$ gives $f(0) = 3$, so the intercept is the point (0, 3).

They could also be expecting the $x$-intercepts, in which case we set $f(x)=0$ and solve for $x$. However, we have

$x^4 - 2x^2 + 3 = \left(x^2 - 1\right)^2 + 2$

and

$x^2-1\ge-1 \implies (x^2-1)^2 \ge (-1)^2 = 1 \implies x^4-2x^2+3 \ge 2$

so there are no $x$-intercepts to worry about.

Relative minima/maxima

Check the sign of the second derivative at each critical point.

$f''(-1) = 8 > 0 \implies \text{rel. min.}$

$f''(0) = -4 < 0 \implies \text{rel. max.}$

$f''(1) = 8 > 0 \implies \text{rel. min.}$

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 $f$ at each of the candidate inflection points found earlier.

$f\left(-\dfrac{\sqrt3}3\right) = \dfrac{22}9$

$f\left(\dfrac{\sqrt3}3\right) = \dfrac{22}9$