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goblinko [34]
2 years ago
6

Given the graph of the function, f(x) , what is the value of f (-2)?

Mathematics
1 answer:
Alexandra [31]2 years ago
3 0

The value of f(-2) is 0

<h3>How to evaluate the function?</h3>

The graph of the function is added as an attachment

The graph is given as:

y = f(x)

When x =-2. we have

y = f(-2)

From the graph, we have:

x = -2 and y = 0

So, we have

f(-2) = 0

Hence, the value of f(-2) is 0

Read more about functions at;

brainly.com/question/2456547

#SPJ1

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Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

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