Find an equation for the line tangent to f(x) = 7 x e^x ? ?

Question

Find an equation for the line tangent to f(x) = 7 x e^x ? ?
Here is question:

Find an equation for the line tangent to the graph of

f(x) = 7 x e^x

at the point (a,f(a)) for a=1.

the answer is:
y =


I do not know how to do this, any help would be great, I keep getting 19.028 and 38.0559 but that is no help it seems

thank you so much!

Answer 1

Slope of the tangent line is the derivative, and since the function is a product of functions, 7x and e^x, use the product rule: (fg)' = fg' + f'g, so

f'(x) = 7e^x + 7xe^x
f'(1) = 7e + 7(1)e = 14e

so at (a,f(a)) = (1,7e), using point-slope form,

y - 7e = 14e(x - 1)
y = 14ex - 14e + 7e
y = 14ex - 7e

You really ought to know that if the question is "find the equation of the tangent," the answer is not going to be just a number like 19 or 38. If you don't know that, you're in WAY over your head and should back up a course or two.

Answer 2

f(x) = 7x*e^x

Find the derivative using the product rule:
f'(x) = 7x*(e^x)' + (e^x)*(7x)' = 7xe^x + 7e^x

Find the slope of the tangent line at x = 1
f'(1) = 7(1)e^1 + 7e^1 = 7e + 7e = 14e

Find the point on the curve:
f(1) = 7(1)e^1 = 7e => x = 1 and y = 7e

Find the equation of the tangent line:
y - 7e = 14e(x - 1)
y - 7e = 14ex - 14e
y = 14ex - 7e

Evaluating e:
y = 38.06x - 19.03

Answer 3

take the derivative to get the slope and then evaluate it is to get the slope at the point...

7e^x + 7xe^x = 7e^x(1+x) = f'(x)
f'(1) = 7e (2) = 14e

the point on the original graph at 1 is:
f(1) = 7(1) e^(1) = 7e

the point is (1,7e) & the slope is 14e

all you need to do now is put it into point-slope form:
y- 7e = 14e (x-1)

Now if you would rather have it in y-intercept:
y = (14e)x -14e +7e

y=(14e)x -7e

Answer 4

via removing: y = t^2 + 2 = [e^(x-7)]^2 + 2 = e^(2x-14) + 2 so dy/dx = 2 e^(2x-14) and at (7, 3) dy/dx = 2e^0 = 2 so the tangent line is (y-3) = 2(x-7) i.e. y = 2x - 11. devoid of removing: dy/dx = dy/dt dt/dx = dy/dt / dx/dt = 2t / (one million/t) = 2t^2 At (7, 3) we've ln t = 0 and subsequently t = one million, so dy/dx = 2. So the tangent line is (y-3) = 2(x-7) i.e. y = 2x - 11.

Answer 5

Yur falt if dis is rong =D. Mite b mine also =D. Chick mi wok.

f(x) = 7e^x (sry din't quite get it)

f'(x) = (0)(e^x) + (7)(e^x)

f'(x) = 7e^x

x = 1

f'(x) = 7e

7e = 7e + b

e = e + b

b = 0

y = 7e*x
______________________________

Ex-q-c muah, how did cha no I rong you rite?

Perhaps, I'd be cor-rect.

No personas say(english) prequntas es 7xe^x. No personas say(english) pregunta es 7e^x. (Hopefully this is correct in spanish)

You will be regretful for give me thumbs down >.<.

(I know my Spanish sucks so don't tell me it.)

Answer 6

http://www.math.hmc.edu/calculus/tutorials/tangent_line/