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Ainat [17]
3 years ago
13

A function is given by the equation h (x)=x 2 squared +6x-3. what is h (3)?

Mathematics
1 answer:
Leno4ka [110]3 years ago
5 0
Your "<span>h (x)=x 2 squared +6x-3" is ambiguous.  If you meant "x squared," then you need only write x^2  OR  "x squared," but NOT "x 2 squared."

I will assume that by  "</span><span>h (x)=x 2 squared +6x-3" you actually meant:

</span><span>h(x)=x^2 +6x-3.  To find h(3), subst. 3 for x:  h(3) = (3)^2 + 6(3) - 3, or

h(3) = 9 + 18 - 3, or h(3) = 24.</span>
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y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the
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The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

For the second condition, we need to find the derivative y' first. In this case, we have:

y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.

Therefore:

y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.

This means that we must solve the following system of equations:

\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.

If we add the equations above, we get:

\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e}  \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.

If we now substitute c_1 = 0 into either of the equations in the system, we get:

c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}

This means that the solution obeying the initial conditions is:

\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

Indeed, we can see that:

y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3

y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,

which do correspond to the desired initial conditions.

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