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Let f(x)=x2+3x−4 .


What is the average rate of change for the quadratic function from x=−3 to x = 8?




Let f(x)=−12(x+2)2+5 .


What is the average rate of change for the quadratic function from x=−3 to x = 1?




Which function grows at the fastest rate for increasing values of x?




g(x)=50x+35


f(x)=5⋅3x


h(x)=6x2+5x+20

Verified Answer :

Answer:

\(\text{Average rate of change of }f(x)=x^2+3x-4 \text{ is 8}\)

\(\text{Average rate of change of }f(x)=-12(x+2)^2+5 \text{ is -24}\)

Option B: \(5.3^x\) grows at the fastest rate for increasing values of x

Step-by-step explanation:

Given the function

\(f(x)=x^2+3x-4\)

we have to find the average rate of change for the quadratic function from x=−3 to x = 8

\(f(-3)=(-3)^2+3(-3)-4=9-9-4=-4\)

\(f(8)=8^2+3(8)-4=64+24-4=84\)

\(\text{Average rate of change=}\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{84-(-4)}{8-(-3)}=\frac{88}{11}=8\)

Now, given function is

\(f(x)=-12(x+2)^2+5\)

we have to find the average rate of change for the quadratic function from x=−3 to x = 1

\(f(-3)=-12(-3+2)^2+5=-12+5=-7\)

\(f(1)=-12(1+2)^2+5=-12(9)+5=-108+5=-103\)

\(\text{Average rate of change=}\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{-103-(-7)}{1-(-3)}=\frac{-96}{4}=-24\)

Now, we have to choose the function which grows at the fastest rate for increasing values of x.

Since we know that quadratic function grows faster than a linear function and an exponential function grows faster than a quadratic function, so the function of an exponential function will have the fastest rate for increasing value of x.

Since an exponential function is in form:

\(y=a.b^x\)

From the given choices we can see that function represented by option A is a linear function, function represented by option B is exponential function and function represented by option C is a quadratic function.

Option B: \(5.3^x\)

∴ option B is the correct choice.

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The average rate of change for the quadratic function from x=−3 to x = 8 is 8.

The function grows at the fastest rate for increasing values of x is \(\rm f(x)=5\times 3^x\)

Given

The given function is;

\(\rm f(x)=x^2+3x-4\)

What is the formula to determine the average value of the function?

The formula to determine the average value of the function is;

\(\rm Average \ value = \dfrac{f(x_1)-f(x_2)}{x+y}\)

The value of the function when x = -3 is;

\(\rm f(x)=x^2+3x-4\\\\\rm f(-3)=(-3)^2+3(-3)-4\\\\f(-3) = 9-9-4\\\\f(-3)=-4\)

The value of the function when x = 8 is;

\(\rm f(x)=x^2+3x-4\\\\\rm f(8)=(8)^2+3(8)-4\\\\f(-3) =64+24-4\\\\f(-3)=84\)

Therefore,

The average rate of change for the quadratic function from x=−3 to x = 8 is;

\(\rm Average \ value = \dfrac{f(x_1)-f(x_2)}{x+y}\\\\\rm Average \ value = \dfrac{84-(-4)}{8-(-3)}\\\\\rm Average \ value = \dfrac{88}{11}\\\\\rm Average \ value = 8\)

The exponential function grows the fastest;

Then,

The function is;

\(\rm f(x)=5\times 3^x\)

Hence, the function grows at the fastest rate for increasing values of x is \(\rm f(x)=5\times 3^x\).

To know more about the Average rate click the link given below.

https://brainly.com/question/18100612

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