What is the average rate of change for the function f(x)=x^2+4 for the domain of -2 to 4
1 answer:
Answer:
2
Step-by-step explanation:
For a quadratic function the average rate of change on an interval is the rate of change at the midpoint of the interval. The rate of change of a function is given by its derivative.
The derivative of f(x) = x^2 is f'(x) = 2x. The midpoint of the interval is (4+(-2))/2 = 1. Then the average rate of change is ...
f'(1) = 2(1) = 2
The average rate of change of f(x) on [-2, 4] is 2.
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<em>Alternate solution</em>
The average rate of change is the slope of the line between the end points of the interval:
m = (y2 -y1)/(x2 -x1)
m = (f(4) -f(-2))/(4 -(-2)) = (20 -8)/(6) = 2
The average rate of change on [-2, 4] is 2.
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The attached graph shows the points on the curve and a line with slope 2 between them. It also shows the various slope calculations.
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Answer:
Step-by-step explanation:
Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).
The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.
We identify the coefficients for the like terms, it comes
a = -2 and 16a = -32 (which is equivalent). So, we can write in .
The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.
And we can write in
Hope this helps.
Do not hesitate if you need further explanation.
Thank you