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)
if you look at the part where the first part connects with the second part:
y = 5 if x < - 2
y = -2x + 1 if -2 ≤ x < 1
we don't have a discontinuity there, so there shouldn't be a dot.
<h3>
</h3><h3>
What is wrong with the graph?</h3>
When we graph over intervals like (a, b) or [a, b] or something like that, we use dots to define the end of the intervals, and to denote that the function ends abruptly or we have a jump.
In this case, you can see that between the end and the second part and the beginning of the third part there is a jump, so the use of dots is correct there, but if you look at the part where the first part connects with the second part:
y = 5 if x < - 2
y = -2x + 1 if -2 ≤ x < 1
we don't have a discontinuity there, so there shouldn't be a dot.
That is the only error with the graph.
If you want to learn more about piecewise functions:
brainly.com/question/3628123
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Answer:
what is the question?
Step-by-step explanation:
Hi!
Use The Percentage Formula To Solve:
385 ÷ 550 * 100 = 70
Mark pays 70% of the original price.
