Question

need help on finding the x-intercept and finding the critical values of x^(5/3)-5x^(2/3) need help on finding the x-intercept and finding the critical values of x^(5/3)-5x^(2/3) @Mathematics

Answer 1

To find the critical values, first you find the derivative and then set it equal to zero.

Answer 2

ok I get (5/3)x^(2/3)-(10/3)x^(-1/3). How would I find the critical value now that I took the derivative and what about the x-intercepts?

Answer 3

Ok the derivative is \[\frac{5(x-2)}{x^{\frac{1}{3}}}\] So set the numerator equal to zero and solve for x.

Answer 4

x = 2, therefore 2 is the critical number

Answer 5

Now to find the x intercepts you do this \[x^{\frac{5}{3}} - 5x^{\frac{2}{3}} = 0\]

Answer 6

is 0 also a critical number ?

Answer 7

and solve for x.

Answer 8

No, 2 is the only critical number.

Answer 9

ok I got x^(5/3)=5^(2/3)

Answer 10

You can automatically say that 0 is a x intercept because 0 raised to the power of anything is 0.

Answer 11

is that the x-intercepts

Answer 12

You can rewrite the problem as \[(x - 5)x^{\frac{2}{3}}=0\] So 5 is a x intercept and 0 is a x intercept.

Answer 13

do I use the 0 and 5 with the 2 from the derivative equation to test the points to see where its increasing/decreasing?

Answer 14

To test whether its increasing or decreasing you first find the derivative and set it equal to zero (we already did that).. This means the tangent line is zero. So you have a interval of \[(-\infty,2]\] and \[[2,\infty)\]

First pick any point in the interval \[(-\infty,2]\] and pug it into the derivative. If you get a negative number then the function is decreasing from the interval (-infinity,2]. If the number is positive then the function is increasing on the same interval.

Next pick any point in the interval \[[2,\infty)\] and do the same thing as you did for the first interval.

Answer 15

how would find the x-intercept of f(x)=(1/4)x^4-2x^2+1=0