We are given a real-valued function
f:[a,∞)⟶[b,∞) defined by
f(x)=2x2−3x+5. This function is a quadratic and is stated to be a bijection.
First, let's find the vertex of the quadratic function, as it will help determine the minimum value of the function, which is crucial for it to be a bijection.
The vertex formula for a quadratic function
ax2+bx+c is given by:
(‌,‌)where
D=b2−4ac. For our function,
a=2,b=−3, and
c=5.
Calculating the vertex:Calculate
x-coordinate of the vertex:
‌=‌Calculate the discriminant
D :
D=(−3)2−4×2×5=9−40=−31Calculate
y-coordinate of the vertex:
‌=‌Thus, the vertex of the quadratic function is
(‌,‌).
Since
f(x)=2x2−3x+5 is a parabola opening upwards (as the coefficient of
x2 is positive), the minimum value of
f(x) occurs at the vertex. Therefore, to achieve a bijective function, the domain should start from the
x-coordinate of the vertex, and the range should start from the
y coordinate of the vertex.
This implies:
a=‌b=‌To find the value of
3a+2b :
3×‌+2×‌=‌+‌=‌+‌=‌=10Thus,
3a+2b=10.