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3 years ago

Given the function f(x) = 20(1.25)^x . what is the average rate of change between f(2) and f(5)

2 answers:

7 0

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)= 20(1.25)^x\implies f(x) = 20\left( \cfrac{5}{4} \right)^x\qquad \begin{cases} x_1=2\\ x_2=5 \end{cases}\implies \cfrac{f(5)-f(2)}{5-2}$

$\bf \cfrac{20\left( \frac{5}{4} \right)^5~~-~~20\left( \frac{5}{4} \right)^2}{3}\implies \cfrac{20\cdot \frac{3125}{1024}~~-~~20\cdot \frac{25}{16}}{3}\implies \cfrac{\frac{5\cdot 3125}{256}-\frac{5\cdot 25}{4}}{3} \\\\\\ \cfrac{\frac{15625}{256}-\frac{125}{4}}{3}\implies \cfrac{\frac{7625}{256}}{3}\implies \cfrac{7625}{256}\cdot \cfrac{1}{3}\implies \cfrac{7625}{768}\implies 9.92838541\overline{6}$

Alex787 [66]3 years ago

3 0

The average rate of change of f(x) on the interval [a,b] is f(b)−f(a)b−a.

We have that a=1254, b=6103100, f(x)=20(54)x.

Thus, f(b)−f(a)b−a=20(54)(6103100)−(20(54)(1254))6103100−(1254)=−58207660913467407226562517167001203595951472642–√5–√4+542101086242752217003726400434970855712890625197922048572373973475376871275743307366424750⋅53100.

Answer: the average rate of change is −58207660913467407226562517167001203595951472642–√5–√4+542101086242752217003726400434970855712890625197922048572373973475376871275743307366424750⋅53100≈550754.870532511

pls put as brainliest

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Given f(x) = (lnx)^3 find the line tangent to f at x = 3

kirill [66]

Explanation

We must the tangent line at x = 3 of the function:

$f(x)=(\ln x)^3.$

The tangent line is given by:

$y=m*(x-h)+k.$

Where:

• m is the slope of the tangent line of f(x) at x = h,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

$f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.$

2) By evaluating the result of f'(x) at x = h = 3, we get:

$m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.$

3) The value of k is:

$k=f(3)=(\ln3)^3$

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

$y=(\ln3)^2*(x-3)+(\ln3)^3.$

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

$y=(\ln3)^2*(x-3)+(\ln3)^3$

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1 year ago

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klasskru [66]

Full working out shown from step 1 to last.

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