find an equation of the curve whose tangent line has a slope of f(x)=8x^(-6/7) given that the point (-1, -6)…

find an equation of the curve whose tangent line has a slope of f(x)=8x^(-6/7) given that the point (-1, -6) is on the curve. the function f(x) satisfying f(x)=8x^(-6/7) and f(-1)= -6 is f(x)= (type an exact answer.)

find an equation of the curve whose tangent line has a slope of f(x)=8x^(-6/7) given that the point (-1, -6) is on the curve. the function f(x) satisfying f(x)=8x^(-6/7) and f(-1)= -6 is f(x)= (type an exact answer.)

Answer

Explanation:

Step1: Integrate (f^{\prime}(x))

We know that if (f^{\prime}(x)=8x^{-\frac{6}{7}}), then by the power - rule of integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). So, (f(x)=\int8x^{-\frac{6}{7}}dx=8\int x^{-\frac{6}{7}}dx). [ \begin{align*} f(x)&=8\times\frac{x^{-\frac{6}{7}+1}}{-\frac{6}{7}+1}+C\ &=8\times\frac{x^{\frac{1}{7}}}{\frac{1}{7}}+C\ &=56x^{\frac{1}{7}}+C \end{align*} ]

Step2: Find the value of (C)

Since (f(-1)=-6), substitute (x = - 1) and (f(x)=-6) into (f(x)=56x^{\frac{1}{7}}+C). [ \begin{align*} -6&=56\times(-1)^{\frac{1}{7}}+C\ -6&=56\times(-1)+C\ -6&=-56 + C\ C&=-6 + 56\ C&=50 \end{align*} ]

Answer:

(f(x)=56x^{\frac{1}{7}}+50)