Question

The function f is such that f(x) = 5 + 6x – x2 for x  3 (a) Express 5 + 6x – x2 in the form p – (x – q) 2 where p and q are constants.Using your answer to part (a), find the range of values of x for which f –1(x) is positive.

Answer 1

Answer:

$\textsf{(a)}\quad f(x)=14-\left(x-3\right)^2$

$\textsf{(a)}\quad x \leq 14$

Step-by-step explanation:

The function f is such that f(x) = 5 + 6x - x² for x ≤ 3.

Part (a)

To express 5 + 6x - x² in the form p - (x - q)² where p and q are constants, we can complete the square.

Rearrange the function to the standard form ax² + bx + c:

$f(x)=-x^2+6x+5$

Factor out the common factor -1:

$f(x)=-(x^2 - 6x - 5)$

Add and subtract the square of half the coefficient of the x-term inside the brackets:

$f(x)=-\left(x^2 - 6x - 5+\left(-\dfrac{6}{2}\right)^2-\left(-\dfrac{6}{2}\right)^2\right)$

$f(x)=-\left(x^2 - 6x - 5+\left(-3\right)^2-\left(-3\right)^2\right)$

$f(x)=-\left(x^2 - 6x - 5+9-9\right)$

$f(x)=-\left(x^2 - 6x+9-14\right)$

Move the 14 outside the brackets so that we have a perfect square trinomial inside the brackets:

$f(x)=-\left(x^2 - 6x+9\right)+14$

Factor the perfect square trinomial:

$f(x)=-\left(x-3\right)^2+14$

Rearrange in the form p - (x - q)²:

$f(x)=14-\left(x-3\right)^2$

Therefore, p = 14 and q = 3.

$\hrulefill$

Part (b)

The domain of the original function is restricted to x ≥ 3. When x = 3, the function is f(3) = 14. As x increases beyond 3, f(x) decreases, so the range of the original function is restricted to f(x) ≤ 14.

The inverse of a function is the reflection of the original function across the line y = x. The domain of the original function corresponds to the range of the inverse function, and vice versa. Therefore, the domain of the inverse function is restricted to x ≤ 14, and its range is restricted to x ≥ 3.

We need to find the range of values of x (domain) for which f⁻¹(x) is positive. Since the range of the inverse function is always positive for the restricted domain, the range of values of x for which f⁻¹(x) is positive is x ≤ 14.

Answer 2

Final answer:

To express f(x) = 5 + 6x - x² in the form p - (x - q)² where p and q are constants, complete the square on the quadratic and linear terms to find p = 14 and q = 3. Then determine the values for which f(x) is greater than 0 to find the range for f⁻¹(x) to be positive, which are between x values of 3 - √14 and 3 + √14.

Explanation:

To express the function f(x) = 5 + 6x – x2 in the form p – (x – q)2, we first complete the square. Completing the square involves reorganizing a quadratic equation into a perfect square trinomial plus or minus a constant.

First, let's rewrite the quadratic term and the linear term: x2 – 6x.

To complete the square, we take half of the linear coefficient (which is 6), square it, and add and subtract this number inside the equation. That number is (6/2)2 = 9. Now add and subtract 9 within the equation:

f(x) = –(x2 – 6x + 9) + 5 + 9

Rewriting –(x2 – 6x + 9) as –(x – 3)2, we have:

f(x) = 14 – (x – 3)2

Where p = 14 and q = 3. We have now expressed the original function in the requested form.

To find the range of values for which f–1(x) is positive, we need to find the values for which the original function, f(x), is greater than 0. Since f(x) is in the form of a downward-opening parabola, it will be greater than 0 between its roots where the parabola is above the x-axis.

Therefore:

x – 3 = 0, when x = 3

x – 3 = √14, when x = 3 + √14 or 3 - √14

The function f(x) will be positive between the values of 3 - √14 and 3 + √14. Hence, those are the values of x for which f–1(x) will also be positive.