# Thread: Skipping the basics

1. ## Skipping the basics

I'm having a problem here. I haven't taken a math class in 6 years but was pretty good in math in high school, so my SAT scores allowed me to skip Math 101 and go to Math 110 algebra.

The concepts of the class are pretty easy for me to grasp, but I'm beginning to find my weakness in a few things I should already know!

Could you help me with three questions?

Finding the inverses of functions.

f(x) = x ^ 3/5

so we have

y = x^3/5
x = y^3/5

Solving for Y to find the inverse is my problem. The answer is x^5/3 = y (the inverse) but I'm not seeing how to get this answer.

Also

f(x) = √4-x²

so again

y = √4-x²

x = √4-y²

solve for y.

I know squaring a square root takes it away, but it seems that my lack of the basic algebra skills is creeping up on me.

Final Problem

Evaluating the function

g(x) = x ^4/3

a. g(8)
b. g(t+1)

so for problem a I can do this..

x ^4/3 = 8 ^4/3 = (8^4)^1/3 = ³√4096 = 16

but for B.

x^4/3 = (t+1)^4/3 , then I'm stuck.

Thanks for the help

2. Originally Posted by kboykb
I'm having a problem here. I haven't taken a math class in 6 years but was pretty good in math in high school, so my SAT scores allowed me to skip Math 101 and go to Math 110 algebra.

The concepts of the class are pretty easy for me to grasp, but I'm beginning to find my weakness in a few things I should already know!

Could you help me with three questions?

Finding the inverses of functions.

f(x) = x ^ 3/5

so we have

y = x^3/5
x = y^3/5

Solving for Y to find the inverse is my problem. The answer is x^5/3 = y (the inverse) but I'm not seeing how to get this answer.

Also

f(x) = √4-x²

so again

y = √4-x²

x = √4-y²

solve for y.

I know squaring a square root takes it away, but it seems that my lack of the basic algebra skills is creeping up on me.

Final Problem

Evaluating the function

g(x) = x ^4/3

a. g(8)
b. g(t+1)

so for problem a I can do this..

x ^4/3 = 8 ^4/3 = (8^4)^1/3 = ³√4096 = 16

but for B.

x^4/3 = (t+1)^4/3 , then I'm stuck.

Thanks for the help
$x=y^{3/5}=\sqrt[5]{y^3}$
$x^5=y^3$
$\sqrt[3]{x^5}=y$
$x^{5/3}=y$

3. Ahhh, now I'm understanding it a bit better. I'll see if I can plug that into the problem. Thanks

4. I'll give you the first question and that should help me with the second

$f(x) = x^\frac{3}{5}$
Replace f(x) with y: $y=x^\frac{3}{5}$
Interchange y and x: $x=y^\frac{3}{5}$
Solve for y
Step 1: Raise both sides by the power of 5: $x^5=y^3$
Step 2: Take the cube root of both sides: $x^\frac{5}{3}=y$

And that's your answer: $f^|(x)=x^\frac{5}{3}
$

You should be able to do the second one now.

Edit: Beaten by VonNemo19

5. Hmm, okay I'm going to take a stab at this.

x = √4-y²

x = (4-y²)^1/2

x² = 4 - y²

x² - 4 = -y²

-x² + 4 = y²

Hm. x + 2 = y

I feel like I just went in the wrong direction.

6. Originally Posted by kboykb
Hmm, okay I'm going to take a stab at this.

x = √4-y² good

x = (4-y²)^1/2 unneccessary, but if it helps you, great!

x² = 4 - y² good

x² - 4 = -y² good

-x² + 4 = y² nice

Hm. x + 2 = y no. +/-(-x^2+4)^1/2=y

I feel like I just went in the wrong direction.
... $\pm\sqrt{-x^2+4}$

7. Okay, I figured that last step is what would get me. Thanks again for the answer and steps, I understand what I need to do now!