Determine all $x$ such that the function $f(x)=-x^3+2x^2+7x+12$ is increasing.
Antwoord 1 correct
Correct
Antwoord 2 optie
All $x$
Antwoord 2 correct
Fout
Antwoord 3 optie
$-2\frac{1}{3}\leq x \leq 1$
Antwoord 3 correct
Fout
Antwoord 4 optie
The correct answer is not among the other options.
Antwoord 4 correct
Fout
Antwoord 1 optie
$-1 \leq x \leq 2\frac{1}{3}$.
Antwoord 1 feedback
Correct: $f'(x)=-3x^2+4x+7$.
Zeros: $x_1=\dfrac{-4-\sqrt{4^2-4\cdot -3\cdot 7}}{2\cdot -3}=2\frac{1}{3}$ and $x_2=\dfrac{-4+\sqrt{4^2-4\cdot -3\cdot 7}}{2\cdot -3}=-1.$
Via a sign chart of $f'(x)$ (for instance with $f'(-2)=-13$, $f'(0)=7$ and $f'(3)=-8$) we find that $f'(x)\geq 0$ for $-1 \leq x \leq 2\frac{1}{3}$.
Go on.
Zeros: $x_1=\dfrac{-4-\sqrt{4^2-4\cdot -3\cdot 7}}{2\cdot -3}=2\frac{1}{3}$ and $x_2=\dfrac{-4+\sqrt{4^2-4\cdot -3\cdot 7}}{2\cdot -3}=-1.$
Via a sign chart of $f'(x)$ (for instance with $f'(-2)=-13$, $f'(0)=7$ and $f'(3)=-8$) we find that $f'(x)\geq 0$ for $-1 \leq x \leq 2\frac{1}{3}$.
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Wrong: Note that you have to use $-b$ in calculating the zeros of a quadratic equation.
See Extra explanation: zeros.
See Extra explanation: zeros.
Antwoord 4 feedback
Wrong: The correct answer is among them.
Try again.
Try again.