Answer to Question #300396 in Calculus for sysysy

Question

16 a Find the y-intercept of the tangent to f (x) = x In x at the point where: i x = 1 it x = 2 iii x = 3. b Make a conjecture about the y-intercept of the tangent to f (x) = x In x at the point where = a, a > 0. c Prove your conjecture algebraically. 17 Find the axes intercepts of the tangent to y = x2 ex at x = 1. 18 Find the exact area of the shaded triangle. 

Answer 1

16.

a.

$f'(x)=(x\ln x)'=\ln x+x(1/x)=\ln x+1$

The equation of the tangent line at $x=x_0$

$y=f'(x_0)x-f'(x_0)x_0+f(x_0)$

i. $x=1$

$f(1)=1\ln(1)=0$

$f'(1)=\ln (1)+1=1$

The equation of the tangent line at $x=1$

$y=x-1$

$x=0, y=-1$

$y$ -intercept: $(0, -1)$

ii. $x=2$

$f(2)=2\ln(2)$

$f'(2)=\ln (2)+1$

The equation of the tangent line at $x=2$

$y=(\ln(2)+1)x-2\ln (2)-2+2\ln (2)$

$y=(\ln(2)+1)x-2$

$x=0, y=-2$

$y$ -intercept: $(0, -2)$

iii. $x=3$

$f(3)=3\ln(3)$

$f'(3)=\ln (3)+1$

The equation of the tangent line at $x=3$

$y=(\ln(3)+1)x-3\ln (3)-3+3\ln (3)$

$y=(\ln(3)+1)x-3$

$x=0, y=-3$

$y$ -intercept: $(0, -3)$

b.

We see that he $y$-intercept of the tangent to $f (x) = x \ln x$ at the point where $x= a, a > 0$ is $(0, -a).$

c. The equation of the tangent line at $x=a, a>0$

$y=f'(a)x-f'(a)a+f(a)$

Substitute

$y=f'(a)x-f'(a)a+f(a)$

$=(\ln(a)+1)x-a\ln (a)-a+a\ln(a)$

$=(\ln(a)+1)x-a$

When $x=0,$ $y=-a.$

$y$ -intercept: $(0, -a)$

17.

$f'=(x^2e^x)'=2xe^x+x^2e^x$

The equation of the tangent line at $x=x_0$

$y=f'(x_0)x-f'(x_0)x_0+f(x_0)$

$x=1$

$f(1)=e$

$f'(1)=3e$

The equation of the tangent line at $x=1$

$y=3ex-2e$

$x=0, y=-2e$

$y$ -intercept: $(0, -2e)$

$y=0, 0=3ex-2e$

$x=2/3$

$x$ -intercept: $(2/3, 0)$

18.

$f'=(3xe^{x/2})'=3e^{x/2}+\dfrac{3}{2}xe^{x/2}$

The equation of the tangent line at $x=x_0$

$y=f'(x_0)x-f'(x_0)x_0+f(x_0)$

$x=-1$

$f(-1)=-3e^{-1/2}$

$f'(-1)=\dfrac{3}{2}e^{-1/2}$

The equation of the tangent line at $x=-1$

$y=\dfrac{3}{2}e^{-1/2}x-\dfrac{3}{2}e^{-1/2}$

$x=0, y=-\dfrac{3}{2}e^{-1/2}$

$y$ -intercept: $(0, -\dfrac{3}{2}e^{-1/2})$

$y=0, 0=\dfrac{3}{2}e^{-1/2}x-\dfrac{3}{2}e^{-1/2}$

$x=1$

$x$ -intercept: $(1, 0)$

$Area=\frac{1}{2}(1)(\dfrac{3}{2}e^{-1/2})=\dfrac{3}{4}e^{-1/2}({units}^2)$