Answer: \(y=\frac12\ln(2.1x^2-0.44)\); \(0.253\); \(x\gt \sqrt{\frac{22}{105}}\).

Let

\(Y=e^{2y}\quad\text{and}\quad X=x^2.\)

The straight line passes through \((4,7.96)\) and \((2,3.76)\), so its gradient is

\(\frac{7.96-3.76}{4-2}=\frac{4.20}{2}=2.1.\)

Thus

\(Y=2.1X+c.\)

Using \((2,3.76)\),

\(3.76=2.1(2)+c.\)

So

\(c=-0.44.\)

Therefore

\(e^{2y}=2.1x^2-0.44.\)

Taking natural logarithms,

\(2y=\ln(2.1x^2-0.44).\)

Hence

\(y=\frac12\ln(2.1x^2-0.44).\)

When \(x=1\),

\(y=\frac12\ln(2.1-0.44)=\frac12\ln1.66.\)

This gives

\(y=0.253\)

to 3 significant figures.

For the straight-line relationship to correspond to a real value of \(y\), the argument of the logarithm must be positive:

\(2.1x^2-0.44\gt 0.\)

So

\(x^2\gt \frac{0.44}{2.1}=\frac{22}{105}.\)

For positive values of \(x\),

\(x\gt \sqrt{\frac{22}{105}}.\)