Sunday

February 1, 2015

February 1, 2015

Posted by **david** on Tuesday, December 5, 2006 at 1:36am.

what???? i don't get get it.

to begin with, i took the derivative if that helps which is 1/(2*sqrtx) .

f'(x) = 1/(2*sqrtx) is correct

f'(x=1) = 1/2

The problem is to find the value of a constant x=c such that the derivative is twice the value at x=1, or 1.

f'(c) = 1 = 1/(2*sqrt c)

2 sqrt c = 1

sqrt c = 1/2

c = 1/4.

**Answer this Question**

**Related Questions**

Math - Could someone help me with these questions, I don't know question c) and ...

Math~Reinyyy - Could someone help me with these questions, I don't know ...

calculus - The length l of a rectangle is decrasing at a rate of 5 cm/sec while ...

Calculus - for f(x)=-x^2 +1, which value is the largest magnitude? a) average ...

CALC SHOULD BE EASYISH - Let f be defined as follows. F(x)=y=x^2 -7 (a) Find the...

calc - Using Ohm's Law (V=IR), where units are volts, amperes, and ohms, and V ...

AP Calc. AB - 1.For f(x)=sin^2(x) and g(x)=0.5x^2 on the interval [-pi/2,pi/2], ...

AP CALC. AB - Let f(x)=3/x. If the rate of change of f at x = c is three times ...

calculas - x and y are both variables that are differentiable with respect to t...

rate of change - Joe is investigating the rate of change of the function y=cos x...