Certainly! Let’s break down each problem and solve it step by step.
### Problem 1: [tex]\( f(t) = 2t - 3; \, f(t^2) = 15 \)[/tex]
We need to find the value of [tex]\( t \)[/tex] such that [tex]\( f(t^2) = 15 \)[/tex].
Given the function [tex]\( f(t) = 2t - 3 \)[/tex]:
[tex]\[ f(t^2) = 2(t^2) - 3 = 15 \][/tex]
Let's solve the equation:
[tex]\[ 2t^2 - 3 = 15 \][/tex]
[tex]\[ 2t^2 = 18 \][/tex]
[tex]\[ t^2 = 9 \][/tex]
Therefore:
[tex]\[ t = \pm 3 \][/tex]
But we only need the value for [tex]\( t^2 \)[/tex]:
[tex]\[ t^2 = 9 \][/tex]
### Problem 2: [tex]\( g(x) = \sqrt{5x - 13}; \, g(9) = ? \)[/tex]
We need to find [tex]\( g(9) \)[/tex].
Given the function [tex]\( g(x) = \sqrt{5x - 13} \)[/tex]:
[tex]\[ g(9) = \sqrt{5 \cdot 9 - 13} = \sqrt{45 - 13} = \sqrt{32} \][/tex]
Thus:
[tex]\[ g(9) = 5.656854249492381 \][/tex]
### Problem 3: [tex]\( f(x) = x^2 - 3x + 5; \, f(3x - 1) = ? \)[/tex]
We need to evaluate [tex]\( f(3x - 1) \)[/tex].
Given the function [tex]\( f(x) = x^2 - 3x + 5 \)[/tex]:
Let [tex]\( y = 3x - 1 \)[/tex]. Substitute [tex]\( y \)[/tex] into the function:
[tex]\[ f(3x - 1) = (3x - 1)^2 - 3(3x - 1) + 5 \][/tex]
First, expand the terms:
[tex]\[ (3x - 1)^2 = 9x^2 - 6x + 1 \][/tex]
[tex]\[ -3(3x - 1) = -9x + 3 \][/tex]
Combine all terms:
[tex]\[ 9x^2 - 6x + 1 - 9x + 3 + 5 = 9x^2 - 15x + 9 \][/tex]
Therefore:
[tex]\[ f(3x - 1) = 9x^2 - 15x + 9 \][/tex]
### Problem 4: [tex]\( g(x) = 3x; \, g\left(\frac{4}{3}\right) = ? \)[/tex]
We need to find [tex]\( g\left(\frac{4}{3}\right) \)[/tex].
Given the function [tex]\( g(x) = 3x \)[/tex]:
[tex]\[ g\left(\frac{4}{3}\right) = 3 \cdot \frac{4}{3} = 4 \][/tex]
Thus:
[tex]\[ g\left(\frac{4}{3}\right) = 4.0 \][/tex]
### Problem 5: [tex]\( f(x) = \frac{5x - 7}{3x - 2}; \, f(2x) = ? \)[/tex]
We need to evaluate [tex]\( f(2x) \)[/tex].
Given the function [tex]\( f(x) = \frac{5x - 7}{3x - 2} \)[/tex]:
Substitute [tex]\( 2x \)[/tex] into the function:
[tex]\[ f(2x) = \frac{5(2x) - 7}{3(2x) - 2} = \frac{10x - 7}{6x - 2} \][/tex]
Therefore:
[tex]\[ f(2x) = \frac{10x - 7}{6x - 2} \][/tex]
In summary, the solutions to the problems are:
1. [tex]\( t^2 = 9 \)[/tex]
2. [tex]\( g(9) = 5.656854249492381 \)[/tex]
3. [tex]\( f(3x - 1) = 9x^2 - 15x + 9 \)[/tex]
4. [tex]\( g\left(\frac{4}{3}\right) = 4.0 \)[/tex]
5. [tex]\( f(2x) = \frac{10x - 7}{6x - 2} \)[/tex]
