Find the inverse of f(x)=2x+1

Mathematics

2 answers:

satela [25.4K]3 years ago

5 0

Answer:

$h(x) = \frac{x-1}{2}$

Step-by-step explanation:

The inverse function is the function which when plugged in to the original one yields the argument (x).

Knowing this we do the following, let h(x) be the inverse function of f:

$f(x)=2x+1 \ \ \ \ f(h(x)) = 2(h(x)) + 1 = x\\h(x) = \frac{x-1}{2}\\Now \ we \ verify \ if \ it \ yields \ x.\\f(h(x)) = f(\frac{x-1}{2}) = 2(\frac{x-1}{2}) + 1 = x -1 + 1 = x\\Thus \ h(x) \ is \ the \ inverse\ function \ of \ f$

Lilit [14]3 years ago

3 0

Answer:

y=0.5(x-1)

Step-by-step explanation:

To find the inverse of a function, you have to substitute yfor x in the equation and simplify to get a normal equation again. In this case, f(x) is y.

y=2x+1

x=2y+1

2y+1=x

2y=x−1

y=0.5(x−1)

So there you have it.

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What is the equation of the quadratic graph with a focus of (-4,17/8) and a directrix of y=15/8

Temka [501]

So hmm notice the graph below

based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"

$\bf \begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}$

now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be $\bf \cfrac{17}{8}-\cfrac{15}{8}=\cfrac{2}{8}\iff \cfrac{1}{4}$

well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either

what's the "p" distance? well, we just found it, is just 1/8

so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8

and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic