Question

Find f f'(x)=1+3sqrt(x) f(4)=25

Answer 1

To find the function f given its derivative f'(x) = 1 + 3sqrt(x) and the condition f(4) = 25, we integrate the derivative and use the condition to solve for the integration constant, resulting in f(x) = x + 2sqrt(x)^2 + 13.

Step-by-step explanation:

The question involves finding a function f given its derivative f'(x) = 1 + 3√x and a condition on the function f(4) = 25. To solve this, we integrate the derivative to find the original function. The solution involves first integrating the derivative:

∫ (1 + 3√x) dx = ∫ dx + 3∫ √x dx = x + 2√x2 + C,

where C is the constant of integration. We use the condition f(4) = 25 to solve for C.

f(4) = 4 + 2√(4)2 + C = 4 + 2(2)2 + C = 4 + 2(4) + C = 4 + 8 + C,

which simplifies to 25 = 12 + C, so C = 13. Therefore, our function is:

f(x) = x + 2√x2 + 13.

Answer 2

It appears that you have a derivative and want to integrate...

dy/dx=1+3√x

y(x)=x+(2/3)x^(3/2)+C you are given that f(4)=25 so we can solve for the constant of integration...

y(4)=25=4+16/3+C

21=16/3+C

(63-16)/3=C

47/3=C so

f(x)=x+(2/3)x^(3/2)+47/3

f(x)=(3x+2x^(3/2)+47)/3