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-\frac13\right) = 12 \left(x - \frac1{\sqrt3}\right) \left(x + \frac1{\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(-\frac{\sqrt3}3\right) = \frac{22}9`

`f\left(\frac{\sqrt3}3\right) = \frac{22}9`