What is the equation of the line tangent to #f(x)= x^5-e^x # at #x=2#?

Question

What is the equation of the line tangent to #f(x)=  x^5-e^x   # at #x=2#?

Isabella Ackerman · Accepted Answer

#y=(80-e^2)*x + e^2-128#

exactly or

#y=72.61 x-120.61#

approximately.

You need to take the derivative and the evaluate it at x=2. We have #f(x)=x^5 -e^x#. We take #{d f(x)}/dx = {d }/dx ( x^5 - e^x) = {d }/dx ( x^5) - {d }/dx ( e^x) # using the Linearity of differentiation. Now we apply the rules for taking the derivative of powers and exponential functions, #D_x x^n = n x^{n-1}# and #D_x e^x =e^x# (isn't Euler's notion so compact?). #{d f(x)}/dx = 5*x^4 - e^x# We need to evaluate this at #x=2#, because this is the slope of the tangent line at x=2. #m=5*2^4 - e^2=80-e^2~= 72.61#. To find the equation of this line we need to know a point on the line, the only point we know is that the line touches the function at #x=2#, so #(2,(f(2))#, is on the line. We need #f(2)#, #f(2)=2^5 -e^2=32-e^2~=24.61#. Recall the equation of the line, #y=m*x+b#, so we have #32-e^2 = (80-e^2)*2+b=160-2*e^2+b#. Rearranging we get #b=32-e^2-160+2e^2=e^2-128~=-120.61#. Putting everything back into the line we have, #y=(80-e^2)*x + e^2-128# exactly or #y=72.61 x-120.61# approximately.

Charles Anctil · Answer

undefined The equation of the line tangent to f(x) = x^5 - e^x at x = 2 is y = 32x - e^2 - 62.

Charles Anctil · Answer

undefined To find the equation of the tangent line to the function $ f(x) = x^5 - e^x $ at $ x = 2 $, follow these steps:

1. Find the slope of the tangent line by taking the derivative of the function $ f(x) $ with respect to $ x $ and evaluating it at $ x = 2 $.

$$ f'(x) = 5x^4 - e^x $$

2. Evaluate $ f'(2) $ to find the slope of the tangent line.

$$ f'(2) = 5(2)^4 - e^2 $$

3. Substitute $ x = 2 $ into the original function $ f(x) $ to find the $ y $-coordinate of the point where the tangent line touches the curve.

$$ f(2) = (2)^5 - e^2 $$

4. Use the point-slope form of the equation of a line to write the equation of the tangent line.

$$ y - f(2) = f'(2)(x - 2) $$

5. Simplify the equation to obtain the final equation of the tangent line.

$$ y - [(2)^5 - e^2] = [5(2)^4 - e^2](x - 2) $$

$$ y - [32 - e^2] = [80 - e^2](x - 2) $$

$$ y = (80 - e^2)x - 128 + e^2 $$