Functions

1. Substitute 3 for every x:
f(3) = 3² − 4(3) + 7
= 9 − 12 + 7
= 4.

2. Substitute a − 1 for every x:
g(a − 1) = 2(a − 1)² − 3(a − 1) + 5
= 2(a² − 2a + 1) − 3a + 3 + 5
= 2a² − 4a + 2 − 3a + 3 + 5
= 2a² − 7a + 10.

3. Substitute −2 for every x:
f(−2) = 3(−2)² − 2(−2) + 1
= 3(4) + 4 + 1
= 17.

4. Substitute t + 1 for every x:
h(t + 1) = (t + 1)² + 2(t + 1) − 3
= t² + 2t + 1 + 2t + 2 − 3
= t² + 4t.

1. Complete the square:
h(x) = x² − 4x + 7
= (x − 2)² − 4 + 7
= (x − 2)² + 3.
The turning point is at x = 2, inside the domain, so the minimum is 3. Check both endpoints:
h(−1) = (−3)² + 3 = 12
h(4) = 2² + 3 = 7.
The maximum is 12, so 3 ≤ h(x) ≤ 12.
Visual check: The solid curve shows only the given domain. T is the minimum; compare the two solid endpoints for the maximum.
A: (−1, 12); turning point T: (2, 3); B: (4, 7).

2. Factor out the negative sign first, then complete the square:
k(x) = −(x² − 4x) + 1
= −[(x − 2)² − 4] + 1
= −(x − 2)² + 5.
The turning point is at x = 2, inside the domain, so the maximum is 5. Check both endpoints:
k(0) = −(−2)² + 5 = 1
k(3) = −1² + 5 = 4.
The minimum is 1, so 1 ≤ k(x) ≤ 5.
Visual check: The turning point T gives the maximum; compare the solid endpoints for the minimum.
A: (0, 1); turning point T: (2, 5); B: (3, 4).

3. Complete the square:
f(x) = (x + 1)² − 1 − 8
= (x + 1)² − 9.
The turning point x = −1 is in the domain, so the minimum is −9.
Check both endpoints:
f(−3) = (−2)² − 9 = −5
f(2) = 3² − 9 = 0.
The maximum is 0, so −9 ≤ f(x) ≤ 0.

4. Factor out −1 from the first two terms, then complete the square:
f(x) = −(x² − 6x) − 5
= −[(x − 3)² − 9] − 5
= −(x − 3)² + 4.
The turning point is at x = 3, which is outside the domain. For 4 ≤ x ≤ 6, the graph decreases as x moves to the right. Check its endpoints:
f(4) = −1 + 4 = 3
f(6) = −9 + 4 = −5.
So −5 ≤ f(x) ≤ 3.
Visual check: T is outside the allowed domain. The dashed part is excluded, so use the solid curve from A to B.
A: (4, 3); turning point T: (3, 4), outside the domain; B: (6, −5).

1. The denominator cannot be zero:
x + 2 ≠ 0
x ≠ −2   (subtract 2).
Rewrite the numerator: 3x − 1 = 3(x + 2) − 7.
Substitute into the original fraction, then split and cancel the common factor:
f(x) = $\frac{3(x+2)-7}{x+2}$
= 3 − $\frac{7}{x+2}$.
The fraction is never zero, so f(x) ≠ 3.
To check the other outputs, let y = 3 − $\frac{7}{x+2}$:
y − 3 = −$\frac{7}{x+2}$
(y − 3)(x + 2) = −7
x + 2 = −$\frac{7}{y-3}$   (divide by y − 3)
x = −2 − $\frac{7}{y-3}$.
For each y ≠ 3, this gives an allowed input. Domain: x ∈ ℝ, x ≠ −2. Range: f(x) ∈ ℝ, f(x) ≠ 3.

2. When x > −2, the denominator is positive, so f(x) < 3. The function approaches 3 from below and decreases without bound near −2. Range: f(x) < 3.

3. The denominator cannot be zero: x + 1 ≠ 0, so x ≠ −1.
Rewrite the numerator: 2x + 7 = 2(x + 1) + 5. Hence
f(x) = 2 + $\frac{5}{x+1}$.
The fraction cannot be zero, so f(x) ≠ 2. To check all other outputs, let y = 2 + $\frac{5}{x+1}$:
y − 2 = $\frac{5}{x+1}$
x + 1 = $\frac{5}{y-2}$
x = $\frac{5}{y-2}$ − 1.
For every y ≠ 2, this gives an allowed input. Thus the range is all real numbers except 2.

4. Because x > 2, the denominator x − 2 is positive. Therefore $\frac{6}{x-2}$ > 0 and g(x) < 4.
To check every output below 4 is possible, let y < 4 and rearrange:
4 − y = $\frac{6}{x-2}$
x − 2 = $\frac{6}{4-y}$
x = 2 + $\frac{6}{4-y}$.
Since 4 − y > 0, this input satisfies x > 2. Thus the range is g(x) < 4.

1. Apply the inside function first:
fg(x) = 2(x² + 3) − 1
= 2x² + 6 − 1 = 2x² + 5.
gf(x) = (2x − 1)² + 3
= 4x² − 4x + 1 + 3
= 4x² − 4x + 4.

2. First find the composite: fg(x) = 2x² + 1.
Now solve:
2x² + 1 = 19
2x² = 18   (subtract 1)
x² = 9   (divide by 2)
x = ±3.

3. For fg, apply g first:
fg(x) = f(3x + 1) = 3x + 1 − 2 = 3x − 1.
For gf, apply f first:
gf(x) = g(x − 2) = 3(x − 2) + 1 = 3x − 5.

4. Apply f first, then g:
gf(x) = (x + 1)².
Set the result equal to 16:
(x + 1)² = 16
x + 1 = 4 or x + 1 = −4
x = 3 or x = −5.

1. Substitute g(x) into f: fg(x) = $\sqrt{x-5}$.
The expression inside the square root must be nonnegative:
x − 5 ≥ 0
x ≥ 5   (add 5 to both sides).

2. Substitute g(x) = x² − 1 into f:
fg(x) = $\frac{1}{(x^2-1)-2}$ = $\frac{1}{x^2-3}$.
The denominator cannot be zero:
x² − 3 ≠ 0
x² ≠ 3   (add 3)
x ≠ ±$\sqrt{3}$.
So the domain is all real numbers except these two values.

3. Apply g first:
fg(x) = $\sqrt{3x-5+2}$ = $\sqrt{3x-3}$.
The square-root expression must be nonnegative:
3x − 3 ≥ 0
3x ≥ 3
x ≥ 1.

4. Substitute the inside function into the denominator:
fg(x) = $\frac{1}{(x^2-5)+1}$
= $\frac{1}{x^2-4}$.
The denominator cannot be zero:
x² − 4 ≠ 0
x² ≠ 4
x ≠ −2 and x ≠ 2.

1. Write y = 4x − 7. Interchange x and y:
x = 4y − 7.
Add 7 to both sides, then divide by 4:
x + 7 = 4y
y = $\frac{x+7}{4}$.
Replace y with f⁻¹(x):
f⁻¹(x) = $\frac{x+7}{4}$, for all real x.

2. Write y = $\frac{2x-1}{x+3}$. Interchange x and y:
x = $\frac{2y-1}{y+3}$.
The original restriction x ≠ −3 now applies to y. Multiply both sides by y + 3:
x(y + 3) = 2y − 1
xy + 3x = 2y − 1.
Collect the y terms and factor:
xy − 2y = −1 − 3x
y(x − 2) = −(3x + 1).
If x = 2, this would give 0 = −7, which is impossible. Thus x ≠ 2. Divide and simplify the signs:
y = $\frac{3x+1}{2-x}$.
So f⁻¹(x) = $\frac{3x+1}{2-x}$.
For every x ≠ 2, y + 3 = $\frac{7}{2-x}$ ≠ 0. The inverse domain is x ∈ ℝ, x ≠ 2.

3. Write y = 5 − 2x and interchange x and y:
x = 5 − 2y.
Subtract 5, then divide by −2:
x − 5 = −2y
y = $\frac{5-x}{2}$.
Therefore f⁻¹(x) = $\frac{5-x}{2}$.

4. Write y = (x − 1)² + 3 and interchange x and y:
x = (y − 1)² + 3.
Subtract 3:
x − 3 = (y − 1)².
The original restriction becomes y ≥ 1 after swapping, so y − 1 ≥ 0. Take the positive square root:
y − 1 = $\sqrt{x-3}$
y = 1 + $\sqrt{x-3}$.
Hence f⁻¹(x) = 1 + $\sqrt{x-3}$. The range of the original function is f(x) ≥ 3, so the inverse domain is x ≥ 3.

1. Swap the coordinates: (5, −2).

2. For f(x) = 2x + 4, the intercepts are (−2, 0) and (0, 4). Reflecting in y = x swaps their coordinates, giving (0, −2) and (4, 0).
For the equation, write y = 2x + 4, then interchange x and y:
x = 2y + 4.
Subtract 4: x − 4 = 2y.
Divide by 2: y = $\frac{x-4}{2}$.
Therefore f⁻¹(x) = $\frac{x-4}{2}$.
Visual check: The blue and gold intercepts swap coordinates across the dashed line y = x.
Corresponding intercepts: (−2, 0) ↔ (0, −2), and (0, 4) ↔ (4, 0).

3. At an intersection with y = x, the two expressions for y are equal:
3x − 4 = x
2x = 4
x = 2, so y = 2.
The point is (2, 2). Swapping its coordinates leaves it unchanged, so it is also on the inverse graph.

4. First find the intercepts of f:
When x = 0, y = 6, giving (0, 6).
When y = 0, −2x + 6 = 0, so x = 3, giving (3, 0).
Swap each pair of coordinates. The inverse intercepts are (6, 0) and (0, 3).
To find the equation, write y = −2x + 6 and interchange variables:
x = −2y + 6
2y = 6 − x
f⁻¹(x) = $\frac{6-x}{2}$.

1. Translate 4 units to the right and 2 units down.

2. In f(x + 3), the +3 is inside the brackets, so translate 3 units left.
The 2 outside the function multiplies the output, so then stretch parallel to the y-axis by scale factor 2.

3. The +2 is inside the brackets, so translate 2 units left.
The −3 is outside the function, so translate 3 units down.

4. One valid order is:
1. The −1 inside the brackets translates the graph 1 unit right.
2. The factor 2 outside gives a stretch parallel to the y-axis with scale factor 2.
3. The negative sign outside gives a reflection in the x-axis.