Consider the function $y(x)=x^4-\frac{3}{2}x^2$. Which of the following statements is true?
The function is convex on the intervals $(-\infty,-\frac{1}{2}]$ and $[\frac{1}{2},\infty)$, and concave on the interval $[-\frac{1}{2},\frac{1}{2}]$.
The function is convex on the interval $(-\infty,\frac{1}{2}]$ and concave on the interval $[\frac{1}{2},\infty)$.
The function is convex on the entire interval.
The function is convex on the interval $(-\infty,0]$, and concave on the interval $[0,\infty)$.
Consider the function $y(x)=x^4-\frac{3}{2}x^2$. Which of the following statements is true?
Antwoord 1 correct
Correct
Antwoord 2 optie
The function is convex on the interval $(-\infty,\frac{1}{2}]$ and concave on the interval $[\frac{1}{2},\infty)$.
Antwoord 2 correct
Fout
Antwoord 3 optie
The function is convex on the entire interval.
Antwoord 3 correct
Fout
Antwoord 4 optie
The function is convex on the interval $(-\infty,0]$, and concave on the interval $[0,\infty)$.
Antwoord 4 correct
Fout
Antwoord 1 optie
The function is convex on the intervals $(-\infty,-\frac{1}{2}]$ and $[\frac{1}{2},\infty)$, and concave on the interval $[-\frac{1}{2},\frac{1}{2}]$.
Antwoord 1 feedback
Correct: From $y''(x)=12x^2-3$ it follows that $y''(x)\geq0$ for every $x\leq-\frac{1}{2}$ and $x\geq\frac{1}{2}$, and that $y''(x)\leq0$ for all other values of $x$.

Go on.
Antwoord 2 feedback
Wrong: Note that the point $x=-\frac{1}{2}$ is also an inflection point.

See inflection point.
Antwoord 3 feedback
Wrong: The function has two inflection points.

See inflection point.
Antwoord 4 feedback
Wrong: $x=0$ is not an inflection point.

See inflection point.