Question

Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point.f '(x) = 4(2x − 9)^3 (5, 11/2)

Answer 1

`f(x)=(1)/(2)(2x-9)^4+5`

Explanation:

Given f'(x) defined below:

`f^(\prime)(x)=4(2x-9)^3_{}`

First, integrate f'(x) to find f(x).

`\int f^(\prime)(x)=\int 4(2x-9)^3dx=4\int (2x-9)^3dx`

Let u = 2x-9

`u=2x-9\implies du=2dx\implies dx=(du)/(2)`

Thus:

`\begin{gathered} f(u)=4\int u^3(du)/(2)=(4)/(2)\int u^3du=(2u^4)/(4)=(1)/(2)u^4+C \\ \implies f(u)=(1)/(2)u^4+C \end{gathered}`

Replace u=2x-9.

`f(x)=(1)/(2)(2x-9)^4+C`

Next, using the point (5,11/2), we find the value of C, the constant of integration.

At (5, 11/2)

`\begin{gathered} x=5,f(x)=(11)/(2) \\ f(x)=(1)/(2)(2x-9)^4+C \\ (11)/(2)=(1)/(2)(2\lbrack5\rbrack-9)^4+C \\ (11)/(2)=(1)/(2)(10-9)^4+C \\ (11)/(2)=(1)/(2)(1)^4+C \\ (11)/(2)=(1)/(2)+C \\ C=(11)/(2)-(1)/(2)=(10)/(2)=5 \end{gathered}`

Therefore, the function f(x) is:

`f(x)=(1)/(2)(2x-9)^4+5`