Solve {l}{f{(x)}={(2x^3+5)}{(3x^4-2)}}{x=1}{g=f{(x)}}{text{Solvefor}htext{where}}{h=g}

$$\left. \begin{array} { l } { f {(x)} = {(2 x ^ {3} + 5)} {(3 x ^ {4} - 2)} }\\ { x = 1 }\\ { g = f {(x)} }\\ { \text{Solve for } h \text{ where} } \\ { h = g } \end{array} \right.$$

Solve for f, x, g, h

$$h=7$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$f\times 1=\left(2\times 1^{3}+5\right)\left(3\times 1^{4}-2\right)$$

Calculate 1 to the power of 3 and get 1.

$$f\times 1=\left(2\times 1+5\right)\left(3\times 1^{4}-2\right)$$

Multiply 2 and 1 to get 2.

$$f\times 1=\left(2+5\right)\left(3\times 1^{4}-2\right)$$

Add 2 and 5 to get 7.

$$f\times 1=7\left(3\times 1^{4}-2\right)$$

Calculate 1 to the power of 4 and get 1.

$$f\times 1=7\left(3\times 1-2\right)$$

Multiply 3 and 1 to get 3.

$$f\times 1=7\left(3-2\right)$$

Subtract 2 from 3 to get 1.

$$f\times 1=7\times 1$$

Multiply 7 and 1 to get 7.

$$f\times 1=7$$

Divide both sides by 1.

$$f=7$$

Consider the third equation. Insert the known values of variables into the equation.

$$g=7\times 1$$

Multiply 7 and 1 to get 7.

$$g=7$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$h=7$$

The system is now solved.

$$f=7 x=1 g=7 h=7$$