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3 years ago

Pls help!!! evaluate the function for x=m/4

Mathematics

1 answer:

EastWind [94]3 years ago

6 0

Plug in m/4 into the equation:

$8 (\frac{m}{4})^{2} - 4(\frac{m}{4})$

Solve the parentheses in the equation:

$\frac{m}{4}^{2} = \frac{m}{4} \times \frac{m}{4} = \frac{m^{2}}{16}$
$8 \times \frac{m^{2}}{16} = \frac{8m^{2}}{16} = \frac{m^{2}}{2}$

$4 \times \frac{m}{4} = \frac{4m}{4} = m$

Your equation should now look like this:

$\frac{m^{2}}{2} - m$

The answer is D.

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$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

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${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

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$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

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