Find an equation to the tangent line for the graph of at the given point integra

Question

Find an equation to the tangent line for the graph of at the given point integral(x)=(2x^4+5)^3, (-1, 49) y=112x-63 y=-112x-63 y=-56x+63 y=-112x+63 y=-56x-63 The ordering and transportation cost C for the components used in manufacturing a product is C=140(420/x^2+x/x+40). X= inequality 1 where C is measured in thousands of dollars and x is the order size in hundred. Find the rate of change of C with respect to x for x=18. Round your answer to two decimal places -28.37 thousand dollars per hundred 19.12 thousand dollars per hundred -18.50 thousand dollars per hundred -20.12 thousand dollars per hundred 21.83 thousand dollars per hundred Find the second derivative of the function. Integral9x)=(3x^3+7)^8 integral^3(x)=75x^3(7+5x^3)(28+75x^3) integral^3(x)=75x^3(7+5x^3)(28-70x^3)

Explanation / Answer

f(x) = (2x4 + 5)2 ; (-1 , 49)

f '(x) = 2(2x4 + 5)2-1 (2(4)x4-1 + 0) ; since d/dx f(g(x)) = f '(g(x)) g '(x) , here f(x) = x2 , g(x) = 2x4 +5; d/dx xn = nxn-1

==> f '(x) = 16x3(2x4 +5)

at x = -1 , f '(x) = f '(-1)

==> f '(-1) = 16(-1)3(2(-1)4 +5) = -16(7) = -112

slope of tangent at (-1 , 49) is f '(-1)

==> equation of line with slope -112 and passing through (-1 , 49) is

y - 49 = -112(x - (-1))

==> y = -112x - 112 + 49

==> y = -112x - 63 is the required equation of tangent.

Option B

2) C = 140(420/x2 + x/(x +40))

differentiating with respect to x

==> C ' = 140[ 420(-2)x-2-1 + [(1)(x +40) - x(1)]/(x +40)2]        since (u/v)' = [u'v - uv']/v2

==> C ' = 140[ -840/x3 + 40/(x + 40)2]

at x = 18 , C ' = 140[ -840/183 + 40/(18 + 40)2]

==> C ' = 140[-35/243 + 10/841]

==> C ' = 140(-0.13214)

==> C ' = -18.4999

==> C ' = -18.5

Hence Option C. $ -18.5 per hundred

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