The function F(x) is a polynomial of degree 3 with coefficients 16, -15, -18, and 6.
The graph of the function will be a cubic curve with two critical points and one inflection point. The graph may have two local minima or maxima depending on the values of the coefficients.
To find the critical points of the function, we can take the derivative of the function and set it equal to zero:
Using the quadratic formula, we can find the solutions for x:
x = (5 ± √(5^2 - 48(-3))) / (2*8) x = (5 ± √(25 + 96)) / 16 x = (5 ± √121) / 16 x = (5 ± 11) / 16 x = 16/16 or x = -6/16
So, the critical points are x = 1 and x = -3/4.
To determine the nature of the critical points, we can use the second derivative test. If F''(x) > 0 at a critical point, then it is a local minimum. If F''(x) < 0, then it is a local maximum.
F''(x) = 96x - 30
At x = 1, F''(1) = 96(1) - 30 = 66, which is positive. So, x = 1 is a local minimum.
At x = -3/4, F''(-3/4) = 96(-3/4) - 30 = -72 - 30 = -102, which is negative. So, x = -3/4 is a local maximum.
Therefore, the function F(x) has a local minimum at x = 1 and a local maximum at x = -3/4.
The function F(x) is a polynomial of degree 3 with coefficients 16, -15, -18, and 6.
The graph of the function will be a cubic curve with two critical points and one inflection point. The graph may have two local minima or maxima depending on the values of the coefficients.
To find the critical points of the function, we can take the derivative of the function and set it equal to zero:
F'(x) = 48x^2 - 30x - 18
Setting F'(x) = 0 and solving for x, we get:
48x^2 - 30x - 18 = 0
6(8x^2 - 5x - 3) = 0
8x^2 - 5x - 3 = 0
Using the quadratic formula, we can find the solutions for x:
x = (5 ± √(5^2 - 48(-3))) / (2*8)
x = (5 ± √(25 + 96)) / 16
x = (5 ± √121) / 16
x = (5 ± 11) / 16
x = 16/16 or x = -6/16
So, the critical points are x = 1 and x = -3/4.
To determine the nature of the critical points, we can use the second derivative test. If F''(x) > 0 at a critical point, then it is a local minimum. If F''(x) < 0, then it is a local maximum.
F''(x) = 96x - 30
At x = 1, F''(1) = 96(1) - 30 = 66, which is positive. So, x = 1 is a local minimum.
At x = -3/4, F''(-3/4) = 96(-3/4) - 30 = -72 - 30 = -102, which is negative. So, x = -3/4 is a local maximum.
Therefore, the function F(x) has a local minimum at x = 1 and a local maximum at x = -3/4.