Question

Please help me on b,c, and d thank you so much for your time! =)
The function meets the following requirements:

i)At x=0, the value of the function is 0, and the slope of the graph of the function is 0.

ii)At x=4, the value of the function is 1, and the slope of the graph of the function is 1.

iii) Between x=0 and x=4, the function is increasing.

a. Let f(x)=ax^2, where (a) is a nonzero constant. Show that it is not possible to find a value for a so that f meets requirement (ii) above.

b. Let g(x)=cx^3-x^2/16, where c is a nonzero constant. Find the value of c so that g meets requirement (ii) above. Show the work that leads to your answer.

c. using the function g and your value of c from part (b), show that g does not meet requirement (iii) above.

d. let h(x)=x^n/k, where k is a nonzero constant and n is a positive integer. find the values of k and n so that h meets requirement (ii) above. show that h also meets requirements (i) and (iii) above.

Answer 1

a) \[1=a(4)^2\]
\[a=\frac{1}{16}\]
\[f(x)=\frac{1}{16}x^2\]
\[f'(x)=\frac{1}{8}x\]
If x = 4, the slope is 2 not 1

Answer 2

So now you try b

Answer 3

how did you get the slope to equal 2? because if you plug 4 into the derivative, shouldn't the slope be 1/2?

Answer 4

Yep. I made a mistake. The conclusion is the same, however.

Answer 5

I understand part a now, yet my part be is not yielding the intended results. Here is my work: c(4^3)-4^2/16=1 thus my c equals 1/32
and so i plug my c back into my original equation g(x)=1/32(x^3)-x^2/16
once i do this i then plug 4 in for my x value and 1 in for my g(x) value and solve the equation to have 1=1, yet when i take the derivative of my g(x) and then substitute in 1 for g'(x) and 4 for x again i get 1=.5 so i am still quite puzzled :/

Answer 6

\[cx^3-\frac{x^2}{16}=1\]
\[c(4^3)-\frac{4^2}{16}=1\]

Answer 7

So I see that you think the whole thing is divided by 16? Why do you think that?

Answer 8

The work you just posted is what i wrote on my piece of paper, i must have typed it wrong on the keyboard, i did not mean to convey the intention of dividing the whole function over 16, rather only (x^2/16), sorry for the misunderstanding :)

Answer 9

Anyway, the real dilemma met is proving that the derivative of g(x), for once 4 is plugged in for x and one is plugged in as g'(x) i do not have one another equal

Answer 10

i rather get 1 equals .5 for some odd reason

Answer 11

Oh crap. I went through the whole thing and then lost it somehow.
Anyway here goes again:

Answer 12

\[g'(x)=\frac{3}{32}x^2-\frac{1}{8}x\]

Answer 13

\[g'(4)=\frac{3}{32}(16)-\frac{1}{8}(4)=\frac{3}{2}-\frac{1}{2}=1\]