Given that f(x) = 2x − 5 and g(x) = x + 3, find (a) f⁻¹(x), (b) f⁻¹(-10), (c) gf(3)

The midpoint formula is $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Given midpoint $(5,2)$, $A(x,9)$, and $B(3,y)$.
For the x-coordinate:

$$5 = \frac{x+3}{2}$$ $$10 = x+3$$ $$x = 10-3$$ $$x = 7$$

For the y-coordinate:

$$2 = \frac{9+y}{2}$$ $$4 = 9+y$$ $$y = 4-9$$ $$y = -5$$

The values are $x=7$ and $y=-5$.

b) The length of line is $8 \text{cm}$. If this is recorded as $8.2 \text{cm}$. Calculate the percentage error.

Absolute error = $| \text{measured value} - \text{actual value} |$

$$\text{Absolute error} = |8.2 \text{cm} - 8 \text{cm}|$$ $$\text{Absolute error} = 0.2 \text{cm}$$

Percentage error = $\frac{\text{Absolute error}}{\text{Actual value}} \times 100\%$

$$\text{Percentage error} = \frac{0.2 \text{cm}}{8 \text{cm}} \times 100\%$$ $$\text{Percentage error} = 0.025 \times 100\%$$ $$\text{Percentage error} = 2.5\%$$

Final Answers for :

a) $x = \boxed{7}$, $y = \boxed{-5}$

b) The percentage error is $\boxed{2.5\%}$

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: Functions

Given that $f(x) = 2x - 5$ and $g(x) = x + 3$, find:

a) $f^{-1}(x)$

Let $y = 2x - 5$.
To find the inverse function, swap $x$ and $y$:

$$x = 2y - 5$$

$$x + 5 = 2y$$ $$y = \frac{x+5}{2}$$

So, $f^{-1}(x) = \frac{x+5}{2}$.

b) $f^{-1}(-10)$

Using $f^{-1}(x) = \frac{x+5}{2}$ from part (a):

$$f^{-1}(-10) = \frac{-10+5}{2}$$ $$f^{-1}(-10) = \frac{-5}{2}$$ $$f^{-1}(-10) = -2.5$$

c) $gf(3)$

Substitute $x=3$ into $f(x) = 2x - 5$:

$$f(3) = 2(3) - 5$$ $$f(3) = 6 - 5$$ $$f(3) = 1$$

Now, calculate $g(f(3))$, which is $g(1)$.
Substitute $x=1$ into $g(x) = x + 3$:

$$g(1) = 1 + 3$$ $$g(1) = 4$$

Final Answers for :

a) $f^{-1}(x) = \boxed{\frac{x+5}{2}}$

b) $f^{-1}(-10) = \boxed{-2.5}$

c) $gf(3) = \boxed{4}$