Identify the open intervals where the function $f ( x ) = 4 x ^ { 2 } + 6 x - 1$ is increasing or decreasing.
A)decreasing: $\left( - \infty , - \frac { 3 } { 4 } \right)$ ; increasing: $\left( - \frac { 3 } { 4 } , \infty \right)$
B)increasing: $\left( - \infty , - \frac { 3 } { 4 } \right)$ ; decreasing: $\left( - \frac { 3 } { 4 } , \infty \right)$
C)increasing on $( - \infty , \infty )$
D)decreasing on $( - \infty , \infty )$
E)none of the above

Question

Quizplus · Accepted Answer

To determine where the function is increasing or decreasing, we first find its derivative: $f'(x) = 8x + 6$. Setting the derivative equal to zero gives the critical point: $8x + 6 = 0 \Rightarrow x = -\frac{3}{4}$. Since the coefficient of $x^2$ in the original function is positive, the function is concave up everywhere, meaning it decreases before and increases after the critical point $-\frac{3}{4}$. Therefore, the function is decreasing on $\left( - \infty , - \frac { 3 } { 4 } \right)$ and increasing on $\left( - \frac { 3 } { 4 } , \infty \right)$.

Identify the Open Intervals Where the Function f(x)=4x2+6x−1f ( x ) = 4 x ^ { 2 } + 6 x - 1f(x)=4x2+6x−1

Identify the Open Intervals Where the Function $f ( x ) = 4 x ^ { 2 } + 6 x - 1$