# Math test review. Any help is very appreciated?

This is not my homework. Its a test review. i have the answers but wondered if anyone could help with any of the problems. Stupid online tutors all want money. Thanks!!

Find f'

f(x)= SR(5x^2 - 3x +1 )

f(x)=(x^2 - 9)^3/2

Find f''

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

f(x) = -4/SR2x+1

Find f(x) = x^2/2x-1

Find the equation of the tangent line to the graph of f(x) = 1/SRx^2+4 at the point (0,1/2)

An object falls from the top of the Empire State Building. Its height(in feet) is given by the equation s(t) = -16t^2 + 1250. Find the velocity and acceleration of the object when it hits the ground.

Find where f(x) = 3x^3 + 12x^2+ 15x is increasing and where it is decreasing along with any relative extrema.

Find where f(x) = x^2/x^2-4 is increasing and where it is decreasing

Find the absolute extrema of f(x) = x^3 -3x^2 on the interval [-1, 3]

Find the absolute extrema of f(x) = (x-1)^2/3 on the interval [-7,2]

### 2 Answers

- Anonymous10 years agoFavorite Answer
>>Stupid online tutors all want money.

No, SMART on-line tutors understandably want money, because they have a skill and a demanded service, which puts them in a position to charge for it. People who give "free lessons" on anything generally suck at what they're doing, which is why they can't charge for it.

What IS extremely stupid and pretentious though is expecting a person such as a tutor to just give out their skills and time for free, let alone to complete strangers.

But since I am an incredibly generous guy, I'll help you with a few of these:

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

On the outside you have the exponent, so the derivative is 3*3(2-x^2)^2, times the derivative of what's inside. Remember that you have to apply the chain rule when you have a function within a function like this. The derivative of 2 - x^2 is 0 - 2x = -2x. So putting it all together gives you

9(2-x^2)^2 * -2x, or -18x(2-x^2)^2.

>>Find the equation of the tangent line to the graph of f(x) =

>>1/SRx^2+4 at the point (0,1/2)

The derivative of a function tells you the slope of a line tangent to it. So take the derivative. Then plug x=0 into the derivative to find the actual slope of the line tangent to the graph at (0,1/2). Now that you have the slope and a point that's on the line (remember a tangent point is part of both the tangent line AND the graph), you can use those to find the equation for the line in the form of y = mx + b. In fact, (0, 1/2) is a y-intercept, so the answer will be y = mx + (1/2), where m is the slope that you got.

>>Find the velocity and acceleration of the object when it hits

>>the ground.

The object hits the ground when the height is 0, which is when s(t) = 0. So set -16t^2 + 1250 equal to 0 and solve for t.

Velocity is change in distance over time, so take the derivative with respect to t to get the equation for velocity. Take the derivative a second time to get acceleration. You found the value of t for when it hits the ground, so plug it into each expression to find the velocity and acceleration at that time.

>>Find where f(x) = 3x^3 + 12x^2+ 15x is increasing and

>>where it is decreasing along with any relative extrema.

A function is increasing when f ' (x) > 0, and decreasing when f ' (x) < 0. So take the derivative, set it to be greater than 0 (for increasing) and solve for x. Do the same for f ' (x) < 0 to find the decreasing range.

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- Anonymous10 years ago
whenever doing derivatives you need to know all your rules. One rule is nx^(n-1). Another thing you need to know is that any root of any number is just that all expression raised to the 1 over the root of the number. For example the cube root of 3 is the same thing as 3^(1/3). Or the square root of x is just x raised to the 1/2.

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