Respuesta :
Answer:
[tex](x,y)=\left(\;\boxed{-\dfrac{\sqrt{10}}{2},-\dfrac{3}{2}}\;\right)\; \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\;\boxed{\dfrac{\sqrt{10}}{2},-\dfrac{3}{2}}\;\right)\; \textsf{(larger $x$-value)}[/tex]
Step-by-step explanation:
To find the points on the graph of the function f(x) = x² - 4 that are closest to the given point (0, -1), we need to minimize the distance between the given point and any point on the graph of the function.
The distance between two points is given by the distance formula:
[tex]\boxed{\begin{array}{l}\underline{\sf Distance \;Formula}\\\\D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\;D\;\textsf{is the distance between two points.} \\\phantom{ww}\bullet\;\;\textsf{$(x_1,y_1)$ and $(x_2,y_2)$ are the two points.}\end{array}}[/tex]
In this case:
- (x₁, y₁) = (0, -1)
- (x₂, y₂) = (x, f(x)) = (x, x² - 4)
So, the distance D(x) between the given point (0, -1) and any point on the graph of f(x) can be expressed as:
[tex]D(x)=\sqrt{(x-0)^2+(x^2-4-(-1))^2}\\\\D(x)=\sqrt{x^2+(x^2-3)^2}\\\\D(x)=\sqrt{x^2+x^4-6x^2+9}\\\\D(x)=\sqrt{x^4-5x^2+9}[/tex]
Now, we need to minimize this distance function to find the points on the graph that are closest to the given point.
To minimize this function, we can take the derivative with respect to x, set it equal to zero, and solve for x.
Differentiate D with respect to x using the chain rule:
[tex]\dfrac{dD}{dx}=\dfrac{d}{dx}\left((x^4-5x^2+9)^{\frac{1}{2}}\right)[/tex]
[tex]\dfrac{dD}{dx}=\dfrac{1}{2}(x^4-5x^2+9)^{-\frac{1}{2}}\cdot \dfrac{d}{dx}\left(x^4-5x^2+9\right)[/tex]
[tex]\dfrac{dD}{dx}=\dfrac{1}{2\sqrt{x^4-5x^2+9}}\cdot (4x^3 -10x)[/tex]
[tex]\dfrac{dD}{dx}=\dfrac{4x^3 -10x}{2\sqrt{x^4-5x^2+9}}[/tex]
[tex]\dfrac{dD}{dx}=\dfrac{2x^3 -5x}{\sqrt{x^4-5x^2+9}}[/tex]
Set the derivative equal to zero and solve for x:
[tex]\dfrac{2x^3 -5x}{\sqrt{x^4-5x^2+9}}=0[/tex]
[tex]2x^3 -5x=0[/tex]
[tex]x(2x^2 -5)=0[/tex]
[tex]\underline{\textsf{Solution 1}}\\\\x=0[/tex]
[tex]\underline{\textsf{Solution 2 \& 3}}\\\\2x^2-5=0\\\\\\x^2=\dfrac{5}{2}\\\\\\x=\pm \sqrt{\dfrac{5}{2}}\\\\\\x=\pm\dfrac{\sqrt{10}}{2}[/tex]
Now evaluate D(x) at each of these x-values to determine the distance between the point and (0, -1).
[tex]D(0)=\sqrt{(0)^4-5(0)^2+9}\\\\D(0)=\sqrt{0-0+9}\\\\D(0)=\sqrt{9}\\\\D(0)=3[/tex]
[tex]D\left(\pm\frac{\sqrt{10}}{2}\right)=\sqrt{\left(\pm\frac{\sqrt{10}}{2}\right)^4-5\left(\pm\frac{\sqrt{10}}{2}\right)^2+9}\\\\\\D\left(\pm\frac{\sqrt{10}}{2}\right)=\sqrt{\frac{25}{4}-\frac{25}{2}+9}\\\\\\D\left(\pm\frac{\sqrt{10}}{2}\right)=\sqrt{\frac{11}{4}}\\\\\\D\left(\pm\frac{\sqrt{10}}{2}\right)=\dfrac{\sqrt{11}}{2}\approx 1.6583[/tex]
So, the x-coordinates of the two points on the graph that are closest to the (0, -1) are:
[tex]x=-\dfrac{\sqrt{10}}{2} \; \;\textsf{and}\;\; x=\dfrac{\sqrt{10}}{2}[/tex]
To find the corresponding y-coordinates, substitute the x-coordinates into f(x):
[tex]f\left(-\frac{\sqrt{10}}{2}\right)=\left(-\frac{\sqrt{10}}{2}\right)^2-4\\\\\\f\left(-\frac{\sqrt{10}}{2}\right)=\dfrac{5}{2}-4\\\\\\f\left(-\frac{\sqrt{10}}{2}\right)=-\dfrac{3}{2}[/tex]
[tex]f\left(\frac{\sqrt{10}}{2}\right)=\left(\frac{\sqrt{10}}{2}\right)^2-4\\\\\\f\left(\frac{\sqrt{10}}{2}\right)=\dfrac{5}{2}-4\\\\\\f\left(\frac{\sqrt{10}}{2}\right)=-\dfrac{3}{2}[/tex]
Therefore, the points on the graph of the function f(x) that are closest to point (0, -1) are:
[tex](x,y)=\left(\;\boxed{-\dfrac{\sqrt{10}}{2},-\dfrac{3}{2}}\;\right)\; \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\;\boxed{\dfrac{\sqrt{10}}{2},-\dfrac{3}{2}}\;\right)\; \textsf{(larger $x$-value)}[/tex]
