## Elementary Linear Algebra 7th Edition

(a) $\langle f, g\rangle= \frac{2}{3}$ (b) $\langle f, f\rangle=\sqrt \frac{2}{3}$ (c) $\| g \| =\sqrt{\langle g, g\rangle}=\sqrt \frac{176}{15}$ (d) $d(f,g)= \sqrt\frac{166}{15}$
$f(x)=-x, \quad g(x)=x^2-x+2.$ (a) $\langle f, g\rangle=\int_{-1}^{1}f(x)g(x) d x=\int_{-1}^{1} (- x^{3}+x^2-2x) d x=\left[-\frac{1}{4}x^{4}+\frac{1}{3}x^{3}- x^{2}\right]_{-1}^{1}=\frac{2}{3}$ (b) $\langle f, f\rangle=\int_{-1}^{1} x^2 d x=\left.\frac{x^3}{3}\right|_{-1} ^{1}=\frac{2}{3} \Longrightarrow\| f \| =\sqrt{\langle f, f\rangle}=\sqrt \frac{2}{3}$ (c) \begin{aligned} \langle g, g\rangle &=\int_{-1}^{1}\left(x^2-x+2\right)^{2} d x \\ &=\int_{-1}^{1} x^{4}-2x^3+5 x^{2}-4x+4 d x \\ &=\frac{1}{5} x^{5}-\frac{1}{2}x^{4}+\frac{5}{3}x^3-2x^2+4\left.x\right|_{-1} ^{1} \\ &=\frac{176}{15}. \end{aligned} Hence, we have $\| g \| =\sqrt{\langle g, g\rangle}=\sqrt \frac{176}{15}$ (d) \begin{aligned}\langle f-g, f-g\rangle &=\int_{-1}^{1}\left(-x^2-2\right)^{2} d x \\ &=\int_{-1}^{1}\left( x^{4}+4x^2+4\right)d x \\ & =\frac{1}{5} x^{5}+\frac{4}{3} x^{3}+\left.4 x\right|_{-1} ^{1} \\ &=\frac{166}{15} \end{aligned}. Hence, we have $d(f,g)=\| f-g \|= \sqrt{\langle f-g, f-g\rangle}=\sqrt\frac{166}{15}$