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EleoNora [17]
3 years ago
12

A is true but f(x) is wrong need help

Mathematics
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

f(x)=-18x^2

Step-by-step explanation:

Given:

1+Integral(f(t)/t^6, t=a..x)=6x^-3

Let's get rid of integral by differentiating both sides.

Using fundamental of calculus and power rule(integration):

0+f(x)/x^6=-18x^-4

Additive Identity property applied:

f(x)/x^6=-18x^-4

Multiply both sides by x^6:

f(x)=-18x^-4×x^6

Power rule (exponents) applied"

f(x)=-18x^2

Check:

1+Integral(-18t^2/t^6, t=a..x)=6x^-3

1+Integral(-18t^-4, t=a..x)=6x^-3

1+(-18t^-3/-3, t=a..x)=6x^-3

1+(6t^-3, t=a..x)=6x^-3

That looks great since those powers are the same on both side after integration.

Plug in limits:

1+(6x^-3-6a^-3)=6x^-3

We need 1-6a^-3=0 so that the equation holds true for all x.

Subtract 1 on both sides:

-6a^-3=-1

Divide both sides by-6:

a^-3=1/6

Raise both sides to -1/3 power:

a=(1/6)^(-1/3)

Negative exponent just refers to reciprocal of our base:

a=6^(1/3)

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