Question

If \(f(t)=\frac{2 t^{2}}{\sqrt{t-1}}\), find \(f(2), f(x+1)\), and \(f(2 x-1)\)

Short answer

\(f(2) = 8\), \(f(x+1) = \frac{2x^2 + 4x + 2}{\sqrt{x}}\), and \(f(2x-1) = \frac{8x^2 - 8x + 2}{\sqrt{2x - 2}}\).

Step-by-step solution

Find f(2)

To find f(2), substitute 2 for t in the function and simplify: \(f(2) = \frac{2(2)^2}{\sqrt{2-1}}\)

Simplify f(2)

Now we will simplify the expression: \(f(2) = \frac{2(4)}{\sqrt{1}} = \frac{8}{1} = 8\)

Find f(x+1)

To find f(x+1), substitute x+1 for t in the function and simplify: \(f(x + 1) = \frac{2(x + 1)^2}{\sqrt{(x + 1) - 1}}\)

Simplify f(x+1)

Now we will simplify the expression: \(f(x + 1) = \frac{2(x^2 + 2x + 1)}{\sqrt{x}} = \frac{2x^2 + 4x + 2}{\sqrt{x}}\)

Find f(2x-1)

To find f(2x-1), substitute 2x-1 for t in the function and simplify: \(f(2x-1) = \frac{2(2x - 1)^2}{\sqrt{(2x - 1) - 1}}\)

Simplify f(2x-1)

Now we will simplify the expression: \(f(2x-1) = \frac{2(4x^2 - 4x + 1)}{\sqrt{2x - 2}} = \frac{8x^2 - 8x + 2}{\sqrt{2x - 2}}\) Now, we have the values for all three expressions: - \(f(2) = 8\) - \(f(x+1) = \frac{2x^2 + 4x + 2}{\sqrt{x}}\) - \(f(2x-1) = \frac{8x^2 - 8x + 2}{\sqrt{2x - 2}}\)