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: $(1, 0)$

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