Differentiation mixed with Line Equation

• September 30th 2009, 07:44 PM
mowgli
Differentiation mixed with Line Equation
the problems goes like this.

f(x) = x^5 + 5e^x , find f'(5)

So I did that and found f'(x) = 5x^4 + 5e^x. Finding f'(5) = 3867

Use this to find the equation of the tangent line to the curve y = x^5 + 5e^x at the point (a,f(a)) when a=5 . The equation of this tangent line can be written in the form y=mx+b where m is: 3867

I found out that part easily, by putting 5 into f'(x)

But the last part of the question is:

and where b is: ?

I have no idea. I tried putting 5 into y = mx+b and adding in the slope but that gives me .2 which is an incorrect answer. Been trying to figure this one out forever!

Thanks,

-MoMo
• September 30th 2009, 07:55 PM
pickslides
To find b in $y=mx+b$

make $y = f(5)$ and $x = 5$

$f(x) = x^5 + 5e^x$

$f(5) = 5^5 + 5e^5 = \dots$

after you have those values together with your gradient solve for b.

To find b in $f(5)=m\times5+b$
• September 30th 2009, 08:00 PM
mowgli
well so f(5) = 5^5 + 5e^5 = 3867

m = 3867 as well

therefore if I do it

f(5) = m * 5 + b

I'm getting 3867 = 3867 * 5 + b

which gives me b = .2

But that is incorrect? Am I missing something? or perhaps my interpretation is wrong..
• September 30th 2009, 08:04 PM
pickslides
Quote:

Originally Posted by mowgli

I'm getting 3867 = 3867 * 5 + b

which gives me b = .2

This is not correct. Look closer.
• September 30th 2009, 08:15 PM
mowgli
i know i will feel really stupid when i see it but I just can't.

f(5) = 3867 and I already discovered that m = 3867 by inputing 5 into f'(x)

I dont see how I can rearrange y = mx+b any other way to discover b.

I feel so frustrated, because I thought it was simple but I dont know why i can't put my head around this.
• September 30th 2009, 08:20 PM
pickslides
$3867 = 3867 \times 5 + b$

$3867 = 19335 + b$

$3867 - 19335 = b$

$b= -15468$
• September 30th 2009, 08:27 PM
mowgli
LOL OMG!!! i've been dividing 3867 by 19355!! Epic Fail T_T

Thank you sooo much!