Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Find the slope of the tangent line to the graph of the given function at the given value of x.

y=-5x^1/2+x^3/2; x=25

Sagot :

olemakpadu
The slope is the differential of the function.

Recall, if y = x^n,  (dy/dx) =  nx^(n-1).

y=-5x^1/2+x^3/2;  x = 25.  To differentiate this, we do it for each term.

(dy/dx) = (1/2)(-5x^(1/2 -1))  +  (3/2)x^(3/2-1)

           =  (-2.5)x^(-0.5)  + 1.5x^(0.5)
            =    (-2.5)/(x^0.5)  +  1.5x^0.5.      Changing -ve power to positive.
                                                              Note x^0.5 = Square rooot (x).

(dy/dx)  =    (-2.5)/(x^0.5)  +  1.5x^0.5.   Note at x = 25.


(dy/dx), x = 25,   =     (-2.5)/(25^0.5)  +  1.5*25^0.5                         
                           =(-2.5)/(5)  +  1.5*5  = -0.5+ 7.5 = 7.

Using also the negative value of square root of 25, which is -5.                          
                            =(-2.5)/(-5)  +  1.5*-5  = 0.5 - 7.5= -7.

Slope at x = 25 is plus or minus 7.
Cheers.

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.