Find the extreme values of the function and where they occur.
-$$y = x ^ { 2 } + 2 x - 3$$
A) The minimum is -4 at x = -1.
B) The minimum is -1 at x = 4.
C) The minimum is 1 at x = -4.
D) The minimum is 1 at x = 4.

Question

Quizplus · Accepted Answer

To find the extreme values, we differentiate the function and set the derivative equal to zero. The derivative of $y = x^2 + 2x - 3$ is $y' = 2x + 2$. Setting $y' = 0$ gives $2x + 2 = 0$, which simplifies to $x = -1$. Substituting $x = -1$ back into the original function gives $y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$. Therefore, the minimum value of the function is -4, which occurs at $x = -1$.

Find the Extreme Values of the Function and Where They y=x2+2x−3y = x ^ { 2 } + 2 x - 3y=x2+2x−3

Find the Extreme Values of the Function and Where They $y = x ^ { 2 } + 2 x - 3$