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

Question

ScanSolve AI Tutor · Accepted Answer

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.

Simplify the function.
$f(x) = 2x^2 - 3x^2 + 1$ $f(x) = -x^2 + 1$
Find the first derivative of the function, $f'(x)$ .
$f'(x) = \frac{d}{dx}(-x^2 + 1)$ $f'(x) = -2x$
Find the critical points by setting the first derivative equal to zero.
$-2x = 0$ $x = 0$
This is the only critical point.
Find the second derivative of the function, $f''(x)$ .
$f''(x) = \frac{d}{dx}(-2x)$ $f''(x) = -2$
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 $x = 0$ .
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 $(0, 1)$ .

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