Respuesta :
f(x) =3x^2+9x-2 has a minimum value. Minimum value of f(x) is [tex]\frac{-35}{4}[/tex]
Solution:
Given, equation is [tex]f(x)=3 x^{2}+9 x-2[/tex]
We have to find whether given equation has maximum or minimum for the given equation.
Now, we know that, f(x) is a quadratic equation and coefficient has [tex]x^2[/tex] is positive, then its graph is upward parabola. Which means that it will have minimum
The minimum value upward parabola will be its vertex.
So, let us convert f(x) into general form. That is,
[tex]f(x)=a(x-h)^{2}+k[/tex]
where a is constant and (h, k) is vertex
[tex]\text { Now }_{,} f(x)=3 x^{2}+9 x-2[/tex]
[tex]\text { Adding and subtracting } \frac{27}{4} \text { for easier calculations }[/tex]
[tex]f(x)=3 x^{2}+9 x+\frac{27}{4}-\frac{27}{4}-2[/tex]
Taking "3" as common from first three terms,
[tex]f(x)=3\left(x^{2}+3 x+\frac{9}{4}\right)-\frac{27}{4}-2[/tex]
Now multiply and divide “2” with “3x” for easier calculations
[tex]f(x)=3\left(x^{2}+2 \times \frac{3}{2} \times x+\left(\frac{3}{2}\right)^{2}\right)-\frac{27+2 \times 4}{4}[/tex]
[tex]\begin{array}{l}{\text { By using }(a+b)^{2}=a^{2}+2 a b+b^{2}, \text { we get }} \\\\ {\left(x^{2}+2 \times \frac{3}{2} \times x+\left(\frac{3}{2}\right)^{2}\right)=\left(x+\frac{3}{2}\right)^{2}}\end{array}[/tex]
[tex]f(x)=3\left(x+\frac{3}{2}\right)^{2}-\frac{35}{4}[/tex]
So, by comparison with general form we get,
[tex]\mathrm{h}=-\frac{3}{2} \text { and } \mathrm{k}=\frac{-35}{4}[/tex]
[tex]\text { Here, }(\mathrm{h}, \mathrm{k})=(\mathrm{x}, \mathrm{f}(\mathrm{x}))=\left(\frac{-3}{2}, \frac{-35}{4}\right)[/tex]
[tex]\text { So, minimum value of } f(x) \text { is } \frac{-35}{4}[/tex]
