Suponga que k es un número real para el cual la gráfica de la función f(x)=x^4+x^3-kx tiene eje de simetrÃa vertical. Se sabe que k se puede expresar como a/b a/b {a;b} son enteros positivos coprimos. Determinar el valor de a+b. Solution.
We have f ( 0 ) = 0 , f ( x 0 = − 1 ) = k k = f ( x 1 ) = x 1 4 + x 1 3 − k x 1 , k 1 / 3 = x 1 Thus f ( − 1 ) = f ( k 1 / 3 ) If f ( x 0 + a x i s ) = f ( − x 0 + a x i s )  By even function definition Then f ( a x i s + d 1 ) = f ( a x i s − d 1 ) = 0 f ( a x i s − d 1 − d 0 ) = f ( − 1 ) f ( a x i s + d 1 + d 0 ) = f ( k 1 / 3 ) f ( − 1 + d 1 + d 0 ) = 0 f ( k 1 / 3 − d 1 − d 0 ) = 0 0 = f ( 0 ) = f ( − 1 + 1 ) = f ( k 1 / 3 − 1 ) Solving for f ( k 1 / 3 − 1 ) = 0 k = 1 8 = a b Hence, a + b = 1 + 8 = 9 \text{We have } f(0)=0,
f(x_0=-1)=k\\
k=f(x_1)=x_1^4+x_1^3-kx_1,k^{1/3}=x_1\\
\text{Thus }f(-1)=f(k^{1/3})\\
\text{If } f(x_0+axis)=f(-x_0+axis)\text{ By even function definition}\\
\text{Then }\\f(axis+d_1)=f(axis-d_1)=0\\
f(axis-d_1-d_0)=f(-1)\\
f(axis+d_1+d_0)=f(k^{1/3})\\
f(-1+d_1+d_0)=0\\
f(k^{1/3}-d_1-d_0)=0\\
0=f(0)=f(-1+1)=f(k^{1/3}-1)\\
\text{Solving for }f(k^{1/3}-1)=0\\
k=\frac{1}{8}=\frac{a}{b}\\
\text{Hence, }a+b=1+8=9 We have f ( 0 ) = 0 , f ( x 0 ​ = − 1 ) = k k = f ( x 1 ​ ) = x 1 4 ​ + x 1 3 ​ − k x 1 ​ , k 1/3 = x 1 ​ Thus f ( − 1 ) = f ( k 1/3 ) If f ( x 0 ​ + a x i s ) = f ( − x 0 ​ + a x i s )  By even function definition Then f ( a x i s + d 1 ​ ) = f ( a x i s − d 1 ​ ) = 0 f ( a x i s − d 1 ​ − d 0 ​ ) = f ( − 1 ) f ( a x i s + d 1 ​ + d 0 ​ ) = f ( k 1/3 ) f ( − 1 + d 1 ​ + d 0 ​ ) = 0 f ( k 1/3 − d 1 ​ − d 0 ​ ) = 0 0 = f ( 0 ) = f ( − 1 + 1 ) = f ( k 1/3 − 1 ) Solving for f ( k 1/3 − 1 ) = 0 k = 8 1 ​ = b a ​ Hence, a + b = 1 + 8 = 9 https://www.geogebra.org/calculator/wnnbsqzj