The relative maximum point of the function f(x) = 2x² -3x² + 1

To find the relative maximum point of the function $f(x) = 2x^2 - 3x^2 + 1$, we first simplify the function and then use calculus. Step 1: Simplify the function. $$ f(x) = 2x^2 - 3x^2 + 1 $$ $$ f(x) = -x^2 + 1 $$ Step 2: Find the first derivative of the function, $f'(x)$. $$ f'(x) = \frac{d}{dx}(-x^2 + 1) $$ $$ f'(x) = -2x $$ Step 3: Find the critical points by setting the first derivative equal to zero. $$ -2x = 0 $$ $$ x = 0 $$ This is the only critical point. Step 4: Find the second derivative of the function, $f''(x)$. $$ f''(x) = \frac{d}{dx}(-2x) $$ $$ f''(x) = -2 $$ Step 5: Use the second derivative test to determine if the critical point is a relative maximum or minimum. Evaluate $f''(x)$ at the critical point $x = 0$: $$ f''(0) = -2 $$ Since $f''(0) = -2 < 0$, the function has a relative maximum at $x = 0$. Step 6: Find the y-coordinate of the relative maximum point by substituting $x = 0$ into the original function. $$ f(0) = -(0)^2 + 1 $$ $$ f(0) = 0 + 1 $$ $$ f(0) = 1 $$ The relative maximum point is $(0, 1)$. The final answer is $\boxed{(0, 1)}$. 3 done, 2 left today. You're making progress.