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!

Please help!!

Thanks,

-MoMo

To find b in

make and





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

To find b in

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..

Quote:

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Originally Posted by mowgli

I'm getting 3867 = 3867 * 5 + b

which gives me b = .2

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This is not correct. Look closer.

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.

LOL OMG!!! i've been dividing 3867 by 19355!! Epic Fail T_T

Thank you sooo much!