Solve f(3/4)=4-2(3/4)^2+(3/4)^4

$$f ( \frac { 3 } { 4 } ) = 4 - 2 ( \frac { 3 } { 4 } ) ^ { 2 } + ( \frac { 3 } { 4 } ) ^ { 4 }$$

Evaluate

$$\frac{817}{256}=3.19140625$$

Solution Steps

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$4-2\times \frac{9}{16}+\left(\frac{3}{4}\right)^{4}$$

Express 2\times \frac{9}{16} as a single fraction.

$$4-\frac{2\times 9}{16}+\left(\frac{3}{4}\right)^{4}$$

Multiply 2 and 9 to get 18.

$$4-\frac{18}{16}+\left(\frac{3}{4}\right)^{4}$$

Reduce the fraction \frac{18}{16} to lowest terms by extracting and canceling out 2.

$$4-\frac{9}{8}+\left(\frac{3}{4}\right)^{4}$$

Convert 4 to fraction \frac{32}{8}.

$$\frac{32}{8}-\frac{9}{8}+\left(\frac{3}{4}\right)^{4}$$

Since \frac{32}{8} and \frac{9}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{32-9}{8}+\left(\frac{3}{4}\right)^{4}$$

Subtract 9 from 32 to get 23.

$$\frac{23}{8}+\left(\frac{3}{4}\right)^{4}$$

Calculate \frac{3}{4} to the power of 4 and get \frac{81}{256}.

$$\frac{23}{8}+\frac{81}{256}$$

Least common multiple of 8 and 256 is 256. Convert \frac{23}{8} and \frac{81}{256} to fractions with denominator 256.

$$\frac{736}{256}+\frac{81}{256}$$

Since \frac{736}{256} and \frac{81}{256} have the same denominator, add them by adding their numerators.

$$\frac{736+81}{256}$$

Add 736 and 81 to get 817.

$$\frac{817}{256}$$

Factor

$$\frac{19 \cdot 43}{2 ^ {8}} = 3\frac{49}{256} = 3.19140625$$