Question

let f be a differentiable function such that f(1)=2 and f′(x)=x2 2cosx 3−−−−−−−−−−−−−√. what is the value of f(4) ?

Answer 1

To find the value of f(4), we integrate the given derivative function and solve for C using the given initial condition. Then, we substitute the value of x = 4 into the equation to find f(4).

Step-by-step explanation:

To find the value of f(4), we need to integrate the given derivative function.

First, we integrate each term separately:

• The antiderivative of x^2 is (1/3)x^3.
• The antiderivative of 2cos(x) is 2sin(x).
• The antiderivative of 3√3 is 3√3x.

Therefore, integrating the derivative function, we get:

f(x) = (1/3)x^3 + 2sin(x) + 3√3x + C,

where C is a constant.

Given that f(1) = 2, we can substitute the values into the equation to solve for C:

2 = (1/3)(1)^3 + 2sin(1) + 3√3(1) + C

Simplifying the equation, we find that C = -2 - (1/3) - 2sin(1) - 3√3.

Finally, we can substitute x = 4 into the equation to find f(4):

f(4) = (1/3)(4)^3 + 2sin(4) + 3√3(4) + C

Substituting the value of C, we can simplify to find the value of f(4).

Answer 2

C) 10.790

Step-by-step explanation:

we are given:

`f(1)=2\\`

`(dy)/(dx) =√(x^2+2cos(x)+3)`

we need to find:

`f(4)=?`

step 1

integrate the function by using a TI-84 calculator (math > 9:fnInt( > input)

`\\\int\limits^4_1 {√(x^2+2cos(x)+3) } \, dx \\\\\\=8.789`

step 2

since we are looking on the interval [1, 4] we need to add our two limiting values together:

`F(4)+F(1)=2+8.789\\\\=10.790`