Question

DIFFERENTIATION
if f(x) = x^3 - 9x^2 + 15x + 3 the relative maxima is ?
A. 5
B. 1
C. 10
D. -22
Ans is C according to the book...plz include working

Q. given that y^3 = x^2 + 1 . Find the change in y if x increases from 1 to 1.1
A.2/15
B.2(^1/3)/30
C.1/15
D.2^(2/3)/30
ans is b.

Answer 1

(1) Using derivatives you can find a candidate for relative maxima or minima,
\[\frac{df(x)}{dx}=3x^2-18x+15\]In order to get a relative maxima or minima, it must be satisfied that,
\[\frac{df(x)}{dx}=0\Rightarrow 3x^2-18x+15=0\Rightarrow x_1=1,x_2=5\]Now you can plot the sign of the derivative or use the second derivative criteria. Using the sign,
\[df(x)/dx>0\ \text{if}| x<1 \\
df(x)/dx<0\ \text{if}| x>1 \\
df(x)/dx<0\ \text{if}| x<5 \\
df(x)/dx>0\ \text{if}| x>5 \\\]The two first lines indicate that in x=1 we have a maximum, while the two last lines indicate we have a minimum. Now we calculate the maximum,
\[f(1)=10\]

(2) For variations in x sufficiently small you have,
\[\Delta y\approx\frac{df(x)}{dx}{\Large|_{x=x_1}}(x-x_1) \]So,
\[\Delta y\approx\frac{2x_1}{3\sqrt[3]{(x_1^2+1)^2}}(x-x_1)\]with x_1=1, and x=1.1, you have,
\[\Delta y=\frac{\sqrt[3]{2}}{30}\]