To find the range of uvx where u(x) = 2x + 3 and v(x) = 1x, we first need to find the expression for uvx. By multiplying these two functions together, we have:
uv(x) = u(x) imes v(x) = (2x + 3) imes (1x) = 2x^2 + 3x
Now, we want to determine the range of the quadratic function 2x^2 + 3x. This is a parabola that opens upwards (since the coefficient of x² is positive), and its range can be found by determining its vertex, which gives us the minimum value of the function.
The x-coordinate of the vertex of a parabola in the form of ax² + bx + c can be found using the formula:
x = -b/(2a)
In our case, a = 2 and b = 3. Plugging these values into the formula gives:
x = -3/(2 * 2) = -3/4
Next, we substitute x = -3/4 back into our function to find the minimum value:
uv(-3/4) = 2(-3/4)² + 3(-3/4)
= 2(9/16) – 9/4
= 18/16 – 36/16 = -18/16 = -9/8
This means that the minimum value of the function is -9/8. Since the parabola opens upwards, the range of uvx is all values greater than or equal to -9/8.
Therefore, the range of uvx is:
[ -9/8, ∞ )