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