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

g(x)=2x-3

h(x)=x-3

find ((f/h))(-1)

asked by guest
on Sep 19, 2024 at 11:49 pm



You asked:

Given the following function definitions: \[\begin{aligned}f\left( x \right) &= {x}^{2} - x - 6\\ g\left( x \right) &= 2 x - 3\\ h\left( x \right) &= x - 3\end{aligned}\] Evaluate \(\frac{f\left( -1 \right)}{h\left( -1 \right)}\).

MathBot Answer:

Assuming the variables \(f\) and \(h\) refer to the functions of the same names. If this is incorrect, please retry your query. Either include arguments for all function calls (i.e. \(f(x) \cdot g(x)\) instead of \(f \cdot g\)) or use different variables in the expression to be evaluated.

$$\begin{aligned}\frac{f\left( -1 \right)}{h\left( -1 \right)}&=1\end{aligned}$$

$$\begin{aligned}\frac{f\left( -1 \right)}{h\left( -1 \right)}&=\frac{\left(-6 - -1 + \left(-1\right)^{2}\right)}{\left(-3 + -1\right)}\\\frac{f\left( -1 \right)}{h\left( -1 \right)}&=\frac{\left(-6 - -1 + 1\right)}{\left(-3 + -1\right)}\\\frac{f\left( -1 \right)}{h\left( -1 \right)}&=\frac{\left(-6 + 2 \cdot 1\right)}{\left(-3 + -1\right)}\\\frac{f\left( -1 \right)}{h\left( -1 \right)}&=\frac{\left(-6 + 2\right)}{\left(-3 + -1\right)}\\\frac{f\left( -1 \right)}{h\left( -1 \right)}&=\frac{\left(-4\right)}{\left(-3 + -1\right)}\\\frac{f\left( -1 \right)}{h\left( -1 \right)}&=1\\\end{aligned}$$

+ Submit Question