### Home > INT3 > Chapter Ch1 > Lesson 1.1.2 > Problem1-25

1-25.

Given $f\left(x\right)=-\frac{2}{3}x+3$ and $g(x)=2x^2-5$, complete parts (a) through (f).

1. Calculate $f(3)$.

Substitute $3$ for each $x$ in $f(x)$.

$f(3)=-\frac{2}{3}(3)+3$

Simplify.

1. Solve $f(x)=-5$.

Substitute $−5$ for $f(x)$.

$-5=-\frac{2}{3}x+3$

Subtract $3$ from both sides.

$-8=-\frac{2}{3}x$

Get the $x$-term alone.

$\left(-\frac{3}{2}\right)(-8)=\left(-\frac{2}{3}x\right)\left(-\frac{3}{2}\right)$

$x=12$

1. Calculate $g(-3)$.

Substitute $−3$ for each $x$ in $g(x)$.

$g(−3)=2(−3)^2−5$

Simplify.

1. Solve $g(x)=-7$.

Substitute $−7$ for $g(x)$.

$−7=2x^2−5$

Solve.

$x^2=−1$

No solution.

1. Solve $g(x)=8$.

Substitute $8$ for $g(x)$.

$8=2x^2−5$

Solve.

$x^2=\frac{13}{2}$

$x=\pm\sqrt\frac{13}{2}$

1. Solve $g(x)=9$.

Substitute $9$ for $g(x)$.

$9=2x^2−5$

Solve.

$x^2=7$

Remember, there should be 2 solutions.

Use the eTool below to solve for each part.
Click the link at right for the full version of the eTool: Int3 1-25 HW eTool