1) \( y=\left(5 x^{2}-3\right)^{2} \)

To analyze the function \( y = (5x^2 - 3)^2 \), we can explore various aspects such as its domain, range, critical points, and behavior. ### 1. Domain The domain of the function is all real numbers since there are no restrictions on \( x \): \[ \text{Domain: } (-\infty, \infty) \] ### 2. Range Since \( (5x^2 - 3)^2 \) is a square of a real number, it is always non-negative. The minimum value occurs when \( 5x^2 - 3 = 0 \): \[ 5x^2 - 3 = 0 \implies 5x^2 = 3 \implies x^2 = \frac{3}{5} \implies x = \pm \sqrt{\frac{3}{5}} \] At these points, the value of \( y \) is: \[ y = (5(\frac{3}{5}) - 3)^2 = (3 - 3)^2 = 0 \] As \( x \) approaches \( \pm \infty \), \( y \) approaches \( \infty \). Therefore, the range is: \[ \text{Range: } [0, \infty) \] ### 3. Critical Points To find the critical points, we need to take the derivative of \( y \) and set it to zero. Using the chain rule: \[ y' = 2(5x^2 - 3) \cdot (10x) = 20x(5x^2 - 3) \] Setting the derivative equal to zero: \[ 20x(5x^2 - 3) = 0 \] This gives us two cases: 1. \( 20x = 0 \) which implies \( x = 0 \) 2. \( 5x^2 - 3 = 0 \) which gives \( x = \pm \sqrt{\frac{3}{5}} \) ### 4. Second Derivative Test To determine the nature of these critical points, we can use the second derivative: \[ y'' = \frac{d}{dx}(20x(5x^2 - 3)) = 20(5x^2 - 3) + 20x(10x) = 20(5x^2 - 3 + 10x^2) = 20(15x^2 - 3) \] Evaluating the second derivative at the critical points: - At \( x = 0 \): \[ y''(0) = 20(15(0)^2 - 3) = 20(-3) = -60 \quad (\text{local maximum}) \] - At \( x = \pm \sqrt{\frac{3}{5}} \): \[ y''\left(\pm \sqrt{\frac{3}{5}}\right) = 20\left(15\left(\frac{3}{5}\right) - 3\right) = 20\left(9 - 3\right) = 120 \quad (\text{local minima}) \] ### 5. Summary - **Domain**: \( (-\infty, \infty) \) - **Range**: \( [0, \infty) \) - **Critical Points**: \( x = 0 \) (local maximum), \( x = \pm \sqrt{\frac{3}{5}} \) (local minima) - **Local Maximum**: \( y(0) = 9 \) - **Local Minima**: \( y\left(\pm \sqrt{\frac{3}{5}}\right) = 0 \) This function is a parabola that opens upwards, with a minimum value of 0 at \( x = \pm \sqrt{\frac{3}{5}} \) and a maximum value of 9 at \( x = 0 \).