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Precalculus : Derivatives

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Example Question #1 : Pre Calculus

The function f(x) is such that

f'(0) = 0

When you take the second derivative of the function f(x) , you obtain

f''(x) = x^2

What can you conclude about the function at x=0 ?

Possible Answers:

The point is an inflection point.

The point is an absolute maximum.

The point is an absolute minimum.

The point is a local maximum.

The point is a local minimum.

Correct answer:

The point is an inflection point.

Explanation:

We have a point at which $f^{'}(x)=0$ . We know from the second derivative test that if the second derivative is negative, the function has a maximum at that point. If the second derivative is positive, the function has a minimum at that point. If the second derivative is zero, the function has an inflection point at that point.

Plug in 0 into the second derivative to obtain

f''(0)=0

So the point is an inflection point.

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Example Question #1 : Pre Calculus

Consider the function

f(x)=x^2-x

Find the maximum of the function on the interval $\left [ 0,1 \right ]$ .

Possible Answers:

-0.5

0.5

Correct answer:

Explanation:

Notice that on the interval $\left [ 0,1 \right ]$ , the term $\ x^{2}$ is always less than or equal to . So the function is largest at the points when x^2=x . This occurs at x=0 and x=1 .

Plugging in either 1 or 0 into the original function f(x) yields the correct answer of 0.

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Example Question #1 : Derivatives

In what -intervals are the relative minimum and relative maximum for the function below?

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

Possible Answers:

$\textup{Relative maximum in}\ \left ( -4,0\right )\textup{and a relative minimum in}\ \left ( 0,1\right )$

$\textup{Relative minimum in}\ \left ( -4,0\right )\textup{and a relative maximum in}\ \left ( 0,1\right )$

$\textup{There are no relative extrema}$

$\textup{Relative maximum in}\ \left ( -1,0\right )\textup{and a relative minimum in}\ \left ( 0,4\right )$

$\textup{Relative minimum in}\ \left ( -1,0\right )\textup{and a relative maximum in}\ \left ( 0,4\right )$

Correct answer:

$\textup{Relative maximum in}\ \left ( -1,0\right )\textup{and a relative minimum in}\ \left ( 0,4\right )$

Explanation:

A cubic function will have at most one relative minimum and one relative maximum. We can determine the zeros be factoring at x=-1,0,4 . From then we only need to determine if the graph is positive or negative in-between the zeros.

The graph is positive between and (plug in x=-0.5 ) and negative between 0 and 4 (plug in x=1 ). This can also be seen from the graph.

Wolframalpha--graph_of_y__x3-3x2-4x_from_x__-5_to_x__5_y__-15_to_y15--2014-12-19_0045

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Example Question #2 : Derivatives

What is the minimum of the function f(x)=x^2-10x+26 ?

Possible Answers:

(5,1)

(7,3)

(4,0)

(6,2)

Correct answer:

(5,1)

Explanation:

The vertex form of a parabola is:

f(x)=a(x-h)^2+k

where {}(h,k) is the vertex of the parabola.

The function for this problem can be simplified into vetex form of a parabola:

f(x)=(x-5)^2+1 ,

with a vertex at (5,1) .

Since the parabola is concave up, the minimum will be at the vertex of the parabola, which is at (5,1) .

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Example Question #1 : Rate Of Change Problems

Find the average rate of change of the function y=2e^x over the interval from x=0 to x=2 .

Possible Answers:

e^2

e^2-1

1-e^2

e(e-1)

Correct answer:

e^2-1

Explanation:

The average rate of change will be found by $\frac{f(2)-f(0)}{2-0}$ .

Here, f(2)=2e^2 , and f(0)=2e^0=2(1)=2 .

Now, we have $\frac{2e^2-2}{2-0}=\frac{2e^2-2}{2}=e^2-1$ .

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Example Question #2 : Rate Of Change Problems

Let a function be defined by f(x)=-ln(x+1) .

Find the average rate of change of the function over [0,4] .

Possible Answers:

$\frac{ln5}{4}$

$-ln\frac{5}{4}$

$ln\frac{-5}{4}$

$-\frac{ln5}{4}$

Correct answer:

$-\frac{ln5}{4}$

Explanation:

We use the average rate of change formula, which gives us $\frac{f(4)-f(0)}{4-0}$ .

Now f(4)=-ln(4+1)=-ln5 , and f(0)=-ln(0+1)=-ln1=0 .

Therefore, the answer becomes $\frac{-ln5-0}{4-0}=-\frac{ln5}{4}$ .

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Example Question #3 : Rate Of Change Problems

Suppose we can model the profit, , in dollars from selling items with the equation P(x)=3x^2-4x+2 .

Find the average rate of change of the profit from x=1 to x=4 .

Possible Answers:

Correct answer:

Explanation:

We need to apply the formula for the average rate of change to our profit equation. Thus we find the average rate of change is $\frac{P(4)-P(1)}{4-1}$ .

Since P(4)=3(4)^2-4(4)+2=3(16)-16+2=48-16+2=34 , and P(1)=3(1)^2-4(1)+2=3(1)-4+2=3-4+2=1 , we find that the average rate of change is $\frac{34-1}{4-1}=\frac{33}{3}=11$ .

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Example Question #1 : Rate Of Change Problems

Let the profit, , (in thousands of dollars) earned from producing items be found by P(x)=2x^2-6x+25 .

Find the average rate of change in profit when production increases from 4 items to 5 items.

Possible Answers:

$\frac{4}{3}$

Correct answer:

Explanation:

Since P(5)=2(5)^2-6(5)+25 , we see that this equals 2(25)-30+25=50-30+25=20+25=45 . Now let's examine P(4) . P(4)=2(4)^2-6(4)+25=2(16)-24+25 which simplifies to .

Therefore the average rate of change formula gives us $\frac{f(5)-f(4)}{5-4}=\frac{45-33}{5-4}=\frac{12}{1}=12$ .

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Example Question #1 : Rate Of Change Problems

Suppose that a customer purchases dog treats based on the sale price , where D(p)=15-p^2+p , where $1\leq p\leq3$ .

Find the average rate of change in demand when the price increases from $2 per treat to $3 per treat.

Possible Answers:

Correct answer:

Explanation:

D(3)=15-9+3=6+3=9

D(2)=15-4+2=13

Thus the average rate of change formula yields $\frac{9-13}{3-2}=\frac{-4}{1}=-4$ .

This implies that the demand drops as the price increases.

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Example Question #1 : Rate Of Change Problems

A college freshman invests $100 in a savings account that pays 5% interest compounded continuously. Thus, the amount saved after years can be calculated by $A(t)=100e^{0.05t}$ .

Find the average rate of change of the amount in the account between t=0 and t=3 , the year the student expects to graduate.

Possible Answers:

$\$38.73$

$-\$5.39$

$\$38.39$

$\$5.39$

Correct answer:

$\$5.39$

Explanation:

$A(3)=100e^{0.05*3}=100e^{0.15}\approx \$116.18$ .

$A(0)=100e^{0.05*0}=100e^{0}=\$100.00$ .

Hence, the average rate of change formula gives us $\frac{A(3)-A(0)}{3-0}=\frac{\$116.18-\$100.00}{3-0}=\frac{\$16.18}{3}\approx \$5.39$ .

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Precalculus : Derivatives

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Pre Calculus

Example Question #1 : Pre Calculus

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #1 : Rate Of Change Problems

Example Question #2 : Rate Of Change Problems

Example Question #3 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

Example Question #1 : Rate Of Change Problems

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