University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

89

Pages:

4
Author:

Herbosambx

Question 1 Find an equation of the tangent line to the given curve at the specified point. 𝑦 = 𝑒 𝑥 𝑥 , ( 1 , 𝑒 ) y= x e x ,(1,e) Answer We can rewrite the original equation in the form: 𝑒 𝑥 ⋅ 𝑥 − 1 e x ⋅x −1 Now, we can differentiate using the Product Rule: ( 𝑒 𝑥 ) ( − 𝑥 − 2 ) + ( 𝑒 𝑥 ) ( 𝑥 − 1 ) (e x )(−x −2 )+(e x )(x −1 ) 𝑒 𝑥 𝑥 2 + 𝑒 𝑥 𝑥 x 2 e x + x e x Simplify: 𝑒 𝑥 𝑥 ( 1 𝑥 + 1 ) x e x ( x 1 +1) In order to find the tangent at our point ( 1 , 𝑒 ) (1,e), we need to plug 1 1 into our derivative. 1 1 + 1 = 𝑒 − 𝑒 = 0 1 1 +1=e−e=0 So 0 0 is our slope at point ( 1 , 𝑒 ) (1,e). To find the normal line, we take the negative reciprocal of our slope: − 1 0 = undefined. − 0 1 =undefined. Which states that our normal to the point ( 1 , 𝑒 ) (1,e) is equal to 𝑥 x. In order to find the tangent, we need to use the Point-Slope formula: 𝑦 − 𝑒 = 0 ( 𝑥 − 1 ) y−e=0(x−1) 𝑦 = 𝑒 y=e Question 2 Find an equation of the line that is tangent to the graph of 𝑓 f and parallel to the given line. Function line: 𝑓 ( 𝑥 ) = 𝑥 2 − 2 𝑥 − 𝑦 + 1 = 0 f(x)=x 2 −2x−y+1=0 Answer Find the slope of the given line slope = 2 slope=2. 2 𝑥 − 𝑦 + 1 = 0 2x−y+1=0 Find the first derivative of 𝑥 x: 𝑓 ′ ( 𝑥 ) = 2 𝑥 f ′ (x)=2x The slope of the given line = the slope of the tangent line because the first line slope = tangent slope. 𝑓 ′ ( 𝑥 ) = 2 𝑥 = 2 f ′ (x)=2x=2 The value of 𝑥 x: 𝑥 = 2 1 = 1 x= 1 2 =1 Substitute in the equation of the curve to find the point of tangency: 𝑓 ( 1 ) = ( 1 ) = 1 𝑦 = 1 f(1)=(1)=1y=1 Question 3 Determine the area of the given region: 𝑦 = 𝑥 + sin ⁡ 𝑥 y=x+sinx Answer We can infer from the graph that the integral's boundaries are between 0 0 and 𝜋 π. Thus, we have: ∫ 0 𝜋 ( 𝑥 + sin ⁡ 𝑥 ) 𝑑 𝑥 = ∫ 0 𝜋 𝑥 𝑑 𝑥 + ∫ 0 𝜋 sin ⁡ 𝑥 𝑑 𝑥 ∫ 0 π (x+sinx)dx=∫ 0 π xdx+∫ 0 π sinxdx = 𝑥 2 2 ∣ 0 𝜋 + ( − cos ⁡ 𝑥 ) ∣ 0 𝜋 = 2 x 2 0 π +(−cosx) 0 π = 𝜋 2 2 − 0 + ( − cos ⁡ 𝜋 + cos ⁡ 0 ) = 2 π 2 −0+(−cosπ+cos0) = 𝜋 2 2 + ( − ( − 1 ) + 1 ) = 2 π 2 +(−(−1)+1) = 𝜋 2 2 + 2 = 2 π 2 +2 Result: 𝜋 2 2 + 2 2 π 2 +2 Question 4 Compose an equation that defines the exponential function with base 𝑎 > 0 a>0. Answer By definition, Base-based exponential function 𝑏 > 0 , 𝑎 ≠ 1 b>0,a  =1 is: 𝑓 ( 𝑥 ) = 𝑏 𝑥 f(x)=b x Exponential function is not defined for 𝑎 < 0 a<0, because that function is not defined for all 𝑥 x. Example: ( − 1 ) 1 / 2 is not defined. (−1) 1/2 is not defined. Answer: 𝑓 ( 𝑥 ) = 𝑏 𝑥 , where 𝑏 > 0 and 𝑎 ≠ 1. f(x)=b x ,where b>0 and a  =1. Question 5 Find an equation of the line described below. Write the equation in slope-intercept form (solved for 𝑦 y), when possible. Through ( 8 , 4 ) (8,4) and ( 4 , 8 ) (4,8). What is the equation of the line? Answer 1 ( 𝑥 1 , 𝑦 1 ) = ( 8 , 4 ) (x 1 ,y 1 )=(8,4) ( 𝑥 2 , 𝑦 2 ) = ( 4 , 8 ) (x 2 ,y 2 )=(4,8) 𝑚 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 m= x 2 −x 1 y 2 −y 1 𝑚 = 8 − 4 4 − 8 = 4 − 4 = − 1 m= 4−8 8−4 = −4 4 =−1 The equation of the line passing through ( 𝑥 1 , 𝑦 1 ) (x 1 ,y 1 ) with slope 𝑚 m is: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) y−y 1 =m(x−x 1 ) 𝑦 − 4 = − 1 ( 𝑥 − 8 ) y−4=−1(x−8) Simplify: 𝑦 = − 𝑥 + 12 y=−x+12