Find the function f(x) described by the given initial value problem.
f′(x)=6x,f(1)=0
6^x dx = 1/ln(6) * 6^x + C
f'(x) = 6^x
so, you have f(x) = 1/ln6 * 6^x + c
f(1) = 0, so
1/ln6 * 6 + c = 0
c = -6/ln6
so, f(x) = 1/ln6 * 6^x - 6/ln6
Find the function f(x) described by the given initial value problem.
f′(x)=8^x , f(2)=1
f'(x) = 8^x
so, you have f(x) = 1/ln8 * 8^x + c
f(2) = 1, so
1/ln8 * 8 + c = 0
do you then take the f(2)=1 and make it
=1/ln8*8^x+c
f(2)=1/ln8*8^x+c=1
c=-1/ln8*-8^+1
c=-9/ln8
To find the function f(x) described by the given initial value problem, we need to solve the differential equation f'(x) = 6x and apply the initial condition f(1) = 0.
To solve the differential equation f'(x) = 6x, we need to integrate both sides with respect to x.
∫ f'(x) dx = ∫ 6x dx
The integral of f'(x) with respect to x is simply f(x), and the integral of 6x with respect to x is (6/2)x^2 = 3x^2.
So, we have:
f(x) = 3x^2 + C
Now, we will use the given initial condition f(1) = 0 to solve for the constant C.
f(1) = 3(1)^2 + C = 3 + C
Since f(1) = 0, we can set 3 + C = 0 and solve for C:
3 + C = 0
C = -3
Therefore, the function f(x) described by the given initial value problem is:
f(x) = 3x^2 - 3